Topic Quiz • 150 questions • 25 seconds per question.
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What is the sum of the first (92) natural numbers?
C. (4278)
\(S_{92}=\frac{92\times93}{2}=4278\) है। बड़े (n) में पहले सम संख्या को आधा करें। / \(S_{92}=\frac{92\times93}{2}=4278\). For large (n), halve the even number first.
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If the sum of the first (n) natural numbers is (4950), what is the value of (n)?
C. (99)
\(S_{99}=\frac{99\times100}{2}=4950\) इसलिए (n=99) है। \(2S_n\) को दो क्रमागत संख्याओं के गुणनफल से मिलाएं। / \(S_{99}=\frac{99\times100}{2}=4950\), so (n=99). Match \(2S_n\) with the product of two consecutive numbers.
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What is the sum of natural numbers from (61) to (125)?
B. (6045)
योग \(S_{125}-S_{60}=7875-1830=6045\) है। बीच की श्रेणी के लिए शुरुआत से पहले का योग घटाएं। / The sum is \(S_{125}-S_{60}=7875-1830=6045\). For a middle range, subtract the sum before the starting point.
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What will be the value of \(S_{150}-S_{140}\)?
C. (1455)
\(S_{150}-S_{140}=141+142+\cdots+150=1455\) है। पास के योगों का अंतर बीच के पदों का योग होता है। / \(S_{150}-S_{140}=141+142+\cdots+150=1455\). The difference of nearby sums is the sum of the terms in between.
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If \(S_n=2080\), what will be the value of \(S_{n+9}\)?
C. (2701)
\(S_{64}=2080\) इसलिए \(S_{73}=2701\) होगा। पहले (n) पहचानें फिर नया सूचकांक लगाएं। / \(S_{64}=2080\), so \(S_{73}=2701\). Identify (n) first and then use the new index.
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What is the value of \(S_{88}+S_{90}-2S_{89}\)?
B. (1)
\(S_{90}=S_{89}+90\) और \(S_{88}=S_{89}-89\) है इसलिए मान (1) है। लगातार योगों के अंतर का उपयोग करें। / \(S_{90}=S_{89}+90\) and \(S_{88}=S_{89}-89\), so the value is (1). Use differences of consecutive sums.
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C. (5430)
अंतर \(S_{120}-S_{60}=7260-1830=5430\) है। तुलना में दोनों योगों को अलग-अलग निकालें। / The difference is \(S_{120}-S_{60}=7260-1830=5430\). In comparison questions, find both sums separately.
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At which (n) will \(S_n\) first become greater than (12000)?
C. (155)
\(S_{154}=11935\) और \(S_{155}=12090\) है इसलिए पहली बार (n=155) पर योग (12000) से अधिक होगा। सीमा प्रश्न में आसपास के मान जांचें। / \(S_{154}=11935\) and \(S_{155}=12090\), so it first exceeds (12000) at (n=155). In boundary questions, check nearby values.
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What is the ratio of \(S_{90}\) to \(S_{45}\)?
A. (91:23)
\(S_{90}:S_{45}=4095:1035=91:23\) है। अनुपात को सबसे बड़े समान गुणक से सरल करें। / \(S_{90}:S_{45}=4095:1035=91:23\). Simplify the ratio by the greatest common factor.
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For which (n) will \(S_n-S_{n-5}=535\)?
C. (109)
अंतर ((n-4)+(n-3)+(n-2)+(n-1)+n=5n-10) है और (5n-10=535) से (n=109)। अंतिम पांच पदों का योग बनाएं। / The difference is ((n-4)+(n-3)+(n-2)+(n-1)+n=5n-10), and (5n-10=535) gives (n=109). Form the sum of the last five terms.
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C. (79)
कुल ब्लॉक \(S_r=3160\) हैं और \(S_{79}=3160\), इसलिए पंक्तियां (79) हैं। ऐसी आकृति में कुल वस्तुएं त्रिभुज संख्या बनाती हैं। / Total blocks are \(S_r=3160\), and \(S_{79}=3160\), so there are (79) rows. Such figures form a triangular number.
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If \(S_a=3003\) and \(S_b=5050\), what will be the value of (b-a)?
C. (23)
\(S_{77}=3003\) और \(S_{100}=5050\), इसलिए (b-a=23)। पहले दोनों सूचकांक पहचानना जरूरी है। / \(S_{77}=3003\) and \(S_{100}=5050\), so (b-a=23). Identifying both indices first is necessary.
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What is the sum of \(86+87+88+\cdots+140\)?
B. (6215)
योग \(S_{140}-S_{85}=9870-3655=6215\) है। (86) से शुरू होने पर (85) तक का योग घटाएं। / The sum is \(S_{140}-S_{85}=9870-3655=6215\). When starting from (86), subtract the sum up to (85).
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If \(S_{m+3}-S_m=282\), what is the value of (m)?
B. (93)
(S_{m+3}-S_m=(m+1)+(m+2)+(m+3)=3m+6) है इसलिए (m=92)। तीन आगे के अंतर में अगले तीन पद जोड़ें। / (S_{m+3}-S_m=(m+1)+(m+2)+(m+3)=3m+6), so (m=92). In a three-step difference, add the next three terms.
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C. (7220)
अंतर \(S_{200}-S_{160}=20100-12880=7220\) है। यह (161) से (200) तक के पदों का योग है। / The difference is \(S_{200}-S_{160}=20100-12880=7220\). It is the sum from (161) to (200).
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At which (n) will \(S_n\) first become greater than (15000)?
B. (173)
\(S_{172}=14878\) और \(S_{173}=15051\) है इसलिए पहली बार (n=173) पर योग (15000) से अधिक होगा। सीमा वाले प्रश्न में पिछले और अगले योग की जांच करें। / \(S_{172}=14878\) and \(S_{173}=15051\), so it first exceeds (15000) at (n=173). In boundary questions, check the previous and next sums.
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What is the value of \(S_{42}+S_{84}+S_{126}\)?
B. (12484)
\(S_{42}=903\), \(S_{84}=3570\) और \(S_{126}=8001\), इसलिए कुल (12474) है। तीनों योग अलग-अलग लिखकर जोड़ें। / \(S_{42}=903\), \(S_{84}=3570\), and \(S_{126}=8001\), so the total is (12474). Write all three sums separately and add.
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If \(S_n=5995\), what will be the value of \(S_{n-8}\)?
C. (5253)
\(S_{109}=5995\) है इसलिए \(S_{101}=5151\) होगा। पहले (n) पहचानें फिर (n-8) लगाएं। / \(S_{109}=5995\), so \(S_{101}=5151\). Identify (n) first and then use (n-8).
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What is the sum of the first (132) natural numbers?
A. (8778)
\(S_{132}=\frac{132\times133}{2}=8778\) है। (132) को आधा करके \(66\times133\) करें। / \(S_{132}=\frac{132\times133}{2}=8778\). Halve (132) and calculate \(66\times133\).
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Find the value of \(S_{81}+S_{108}-S_{54}\).
B. (7722)
\(S_{81}=3321\), \(S_{108}=5886\) और \(S_{54}=1485\), इसलिए मान (7722) है। मिश्रित योगों में हर \(S_n\) अलग लिखें। / \(S_{81}=3321\), \(S_{108}=5886\), and \(S_{54}=1485\), so the value is (7722). In mixed sums, write each \(S_n\) separately.
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If \(S_{4p}=7875\), what is the value of (p)?
B. (31)
\(S_{125}=7875\) है इसलिए (4p=125) से पूर्णांक (p) नहीं मिलता। ऐसे प्रश्नों में सूचकांक की विभाज्यता भी जांचें। / \(S_{125}=7875\), so (4p=125) gives no integer (p). In such questions, check divisibility of the index too.
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What will be the value of \(S_{144}-2S_{72}\)?
A. (5040)
\(S_{144}=10440\) और \(2S_{72}=5256\), इसलिए मान (5184) है। गुणा और घटाव का क्रम ध्यान से रखें। / \(S_{144}=10440\) and \(2S_{72}=5256\), so the value is (5184). Keep the order of multiplication and subtraction clear.
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On a stage, the (r)th row has (6r) lamps. How many lamps will there be in (45) rows?
B. (6210)
कुल दीपक \(6S_{45}=6\times1035=6210\) हैं। समान गुणक हो तो उसे योग के बाहर रखें। / Total lamps are \(6S_{45}=6\times1035=6210\). If there is a common multiplier, keep it outside the sum.
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If \(S_n=7260\), what will be the value of (n)?
C. (120)
\(S_{120}=\frac{120\times121}{2}=7260\) इसलिए (n=120) है। \(2S_n\) को क्रमागत संख्याओं के गुणनफल से मिलाएं। / \(S_{120}=\frac{120\times121}{2}=7260\), so (n=120). Match \(2S_n\) with the product of consecutive numbers.
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The value of \(S_n-S_{n-2}+S_{n-6}-S_{n-8}\) is (312). What is (n)?
A. (80)
व्यंजक ((n-1)+n+(n-7)+(n-6)=4n-14) है और (4n-14=312) से पूर्णांक (n) नहीं मिलता। पहले पदों में बदलकर समता जांचें। / The expression is ((n-1)+n+(n-7)+(n-6)=4n-14), and (4n-14=312) gives no integer (n). Convert into terms and check parity first.
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Which number should be added to \(S_{95}\) to get \(S_{105}\)?
A. (1005)
जरूरी संख्या \(96+97+\cdots+105=1005\) है। बीच के सभी नए पद जोड़ें। / The needed number is \(96+97+\cdots+105=1005\). Add all new terms in between.
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If \(S_u=2346\) and \(S_v=6670\), what is the value of (u+v)?
B. (182)
\(S_{68}=2346\) और \(S_{115}=6670\), इसलिए (u+v=183) है। दोनों त्रिभुज संख्याओं के सूचकांक पहचानें। / \(S_{68}=2346\) and \(S_{115}=6670\), so (u+v=183). Identify the indices of both triangular numbers.
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What is (12.5%) of the sum of the first (144) natural numbers?
B. (1305)
\(S_{144}=10440\) है और \(12.5%=\frac{1}{8}\), इसलिए मान (1305) है। प्रतिशत को भिन्न में बदलना आसान रहता है। / \(S_{144}=10440\), and \(12.5%=\frac{1}{8}\), so the value is (1305). Converting percent to a fraction makes it easier.
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What is the value of \(S_{102}+S_{103}-S_{101}\)?
C. (5554)
\(S_{102}=5253\), \(S_{103}=5356\) और \(S_{101}=5151\), इसलिए मान (5458) है। निकट सूचकांकों में सावधानी से घटाएं। / \(S_{102}=5253\), \(S_{103}=5356\), and \(S_{101}=5151\), so the value is (5458). Subtract carefully with nearby indices.
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If \(S_n-S_{n-7}=1001\), what will be the value of (n)?
C. (142)
अंतर ((n-6)+(n-5)+\cdots+n=7n-21) है और (7n-21=1001) से (n=146) मिलता है। अंतिम सात पदों का योग बनाएं। / The difference is ((n-6)+(n-5)+\cdots+n=7n-21), and (7n-21=1001) gives (n=146). Form the sum of the last seven terms.
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What is the sum of the first (250) natural numbers?
B. (31375)
\(S_{250}=\frac{250\times251}{2}=31375\) है। बड़े (n) में पहले (250) को (125) बनाएं। / \(S_{250}=\frac{250\times251}{2}=31375\). For large (n), first change (250) to (125).
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Find the value of \(S_{180}-S_{170}+S_{20}\).
B. (1975)
(16290-14535+210=1965) है। पहले घटाव करें फिर छोटे योग को जोड़ें। / (16290-14535+210=1965). First subtract and then add the smaller sum.
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If \(S_{2k+3}=3240\), what will be the value of (k)?
D. (40)
\(S_{80}=3240\) इसलिए (2k+3=80) से पूर्णांक (k) नहीं मिलता। सूचकांक में दी गई शर्त को भी जांचें। / \(S_{80}=3240\), so (2k+3=80) gives no integer (k). Check the given index condition too.
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What is the sum of \(151+152+153+\cdots+210\)?
A. (10830)
योग \(S_{210}-S_{150}=22155-11325=10830\) है। (151) से शुरू होने पर (150) तक का योग घटाएं। / The sum is \(S_{210}-S_{150}=22155-11325=10830\). When starting from (151), subtract the sum up to (150).
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If \(S_n=4656\), what is the value of \(S_{n+12}-S_n\)?
A. (1230)
\(S_{96}=4656\) इसलिए अंतर \(97+98+\cdots+108=1230\) है। आगे के (12) पदों का योग अलग से निकालें। / \(S_{96}=4656\), so the difference is \(97+98+\cdots+108=1230\). Find the sum of the next (12) terms separately.
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At which (n) will \(S_n\) first become greater than (30000)?
B. (245)
\(S_{244}=29890\) और \(S_{245}=30135\) है इसलिए पहली बार (n=245) पर योग (30000) से अधिक होगा। आसपास के मान जांचना सबसे सुरक्षित है। / \(S_{244}=29890\) and \(S_{245}=30135\), so it first exceeds (30000) at (n=245). Checking nearby values is the safest method.
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A. (11172)
कुल वाहन \(7S_{56}=7\times1596=11172\) हैं। (7r) जैसे पैटर्न में (7) को योग के बाहर लें। / Total vehicles are \(7S_{56}=7\times1596=11172\). In a pattern like (7r), take (7) outside the sum.
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If \(S_x=8385\), what is the value of (x)?
B. (129)
\(S_{129}=\frac{129\times130}{2}=8385\) इसलिए (x=129) है। \(2S_x\) करके क्रमागत गुणनफल खोजें। / \(S_{129}=\frac{129\times130}{2}=8385\), so (x=129). Use \(2S_x\) to find the consecutive product.
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What is the value of \(S_{48}+S_{96}-S_{32}\)?
B. (5294)
(1176+4656-528=5304) है। हर \(S_n\) का मान पहले लिखकर फिर क्रिया करें। / (1176+4656-528=5304). Write the value of each \(S_n\) first and then perform the operation.
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C. (7525)
अंतर \(S_{175}-S_{125}=15400-7875=7525\) है। यह (126) से (175) तक के पदों का योग है। / The difference is \(S_{175}-S_{125}=15400-7875=7525\). It is the sum from (126) to (175).
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If \(S_n-S_{n-1}=154\) and \(S_m=7503\), what is (n-m)?
B. (31)
पहले (n=154) और \(S_m=7503\) से (m=122), इसलिए (n-m=32)। लगातार योगों का अंतर सूचकांक देता है। / First (n=154), and \(S_m=7503\) gives (m=122), so (n-m=32). The difference of consecutive sums gives the index.
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Which sum is obtained by adding \(133+134+\cdots+140\) to \(S_{132}\)?
C. \(S_{140}\)
\(S_{132}+133+134+\cdots+140=S_{140}\) है। लगातार नए पद जुड़ने पर अंतिम पद नया सूचकांक बनता है। / \(S_{132}+133+134+\cdots+140=S_{140}\). When consecutive new terms are added, the last term becomes the new index.
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If \(S_n=11781\), what is the value of (n)?
C. (153)
\(S_{153}=\frac{153\times154}{2}=11781\) इसलिए (n=153) है। बड़े मानों में \(2S_n\) से पहचान आसान होती है। / \(S_{153}=\frac{153\times154}{2}=11781\), so (n=153). For large values, using \(2S_n\) makes identification easier.
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Find the value of \(S_{37}+S_{74}+S_{111}\).
A. (9775)
\(S_{37}=703\), \(S_{74}=2775\) और \(S_{111}=6216\), इसलिए कुल (9694) है। तीन अलग सूचकांकों को ध्यान से जोड़ें। / \(S_{37}=703\), \(S_{74}=2775\), and \(S_{111}=6216\), so the total is (9694). Add the three different indices carefully.
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The sum of the first (n) natural numbers is (13695). What is the value of (n-30)?
B. (135)
\(S_{165}=13695\) इसलिए (n=165) और (n-30=135)। पहले (n) निकालें फिर पूछा गया मान दें। / \(S_{165}=13695\), so (n=165) and (n-30=135). Find (n) first and then answer what is asked.
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If \(S_{n+3}-S_{n-3}=615\), what will be the value of (n)?
D. (102)
अंतर ((n-2)+(n-1)+n+(n+1)+(n+2)+(n+3)=6n+3) है और (6n+3=615) से (n=102)। बीच के छह पदों का योग बनाएं। / The difference is ((n-2)+(n-1)+n+(n+1)+(n+2)+(n+3)=6n+3), and (6n+3=615) gives (n=102). Form the sum of the six middle terms.
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What is the value of \(S_{300}-S_{285}\)?
B. (4395)
अंतर \(286+287+\cdots+300=4395\) है। बड़े सूचकांकों में भी केवल बीच के पद जोड़ने होते हैं। / The difference is \(286+287+\cdots+300=4395\). Even with large indices, only the terms in between need to be added.
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If \(S_n=20100\), what is the value of (n)?
C. (200)
\(S_{200}=\frac{200\times201}{2}=20100\) इसलिए (n=200) है। \(2S_n\) को क्रमागत गुणनफल से मिलाएं। / \(S_{200}=\frac{200\times201}{2}=20100\), so (n=200). Match \(2S_n\) with a consecutive product.
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What is (4) times the difference between \(S_{96}\) and \(S_{48}\)?
A. (13920)
\(S_{96}-S_{48}=4656-1176=3480\) और \(4\times3480=13920\) है। पहले अंतर निकालें फिर गुणा करें। / \(S_{96}-S_{48}=4656-1176=3480\), and \(4\times3480=13920\). First find the difference and then multiply.
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If \(S_y=15931\), what will be the value of \(S_{y+3}\)?
A. (16435)
\(S_{178}=15931\) है इसलिए \(S_{181}=15931+179+180+181=16471\)। तीन आगे जाने पर अगले तीन पद जोड़ें। / \(S_{178}=15931\), so \(S_{181}=15931+179+180+181=16471\). To move three steps ahead, add the next three terms.
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If (S_n=\frac{n(n+1)}{2}) and \(S_n=6670\), what is the value of (n)?
C. (115)
\(\frac{115\cdot116}{2}=6670\), इसलिए (n=115)। बड़े मानों में विकल्पों को सूत्र में रखकर जांचें। / \(\frac{115\cdot116}{2}=6670\), so (n=115). For large values, check options in the formula.
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What is the sum of natural numbers from (41) to (95)?
A. (3740)
योग \(S_{95}-S_{40}=4560-820=3740\) है। सीमा से पहले तक का योग जरूर घटाएं। / The sum is \(S_{95}-S_{40}=4560-820=3740\). Always subtract the sum before the lower limit.
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If \(S_n-S_{n-5}=490\), what will be the value of (n)?
C. (100)
(S_n-S_{n-5}=(n-4)+(n-3)+(n-2)+(n-1)+n=5n-10)। (5n-10=490) से (n=100)। / (S_n-S_{n-5}=(n-4)+(n-3)+(n-2)+(n-1)+n=5n-10). From (5n-10=490), (n=100).
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If \(S_{n+3}-S_n=222\), what is the value of (n)?
B. (72)
(S_{n+3}-S_n=(n+1)+(n+2)+(n+3)=3n+6)। (3n+6=222) से (n=72)। / (S_{n+3}-S_n=(n+1)+(n+2)+(n+3)=3n+6). From (3n+6=222), (n=72).
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What is the difference between the sums of the first (120) and first (85) natural numbers?
C. (3605)
\(S_{120}-S_{85}=7260-3655=3605\)। यह (86) से (120) तक का योग है। / \(S_{120}-S_{85}=7260-3655=3605\). This is the sum from (86) to (120).
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If \(S_n=5050\), what is the value of \(S_{n+4}-S_n\)?
B. (410)
\(5050=S_{100}\), इसलिए अंतर (101+102+103+104=410) है। अगले पदों को सीधे जोड़ें। / \(5050=S_{100}\), so the difference is (101+102+103+104=410). Add the next terms directly.
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What is the ratio of the sums of the first (64) and first (32) natural numbers?
A. (65:33)
\(S_{64}:S_{32}=2080:528=65:33\)। अनुपात को सरलतम रूप में लिखें। / \(S_{64}:S_{32}=2080:528=65:33\). Write the ratio in simplest form.
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If \(S_n+S_{n-1}=4212\) and \(S_n-S_{n-1}=65\), what is the value of \(S_n\)?
A. (2138.5)
दोनों समीकरण जोड़ने पर \(2S_n=4277\), इसलिए \(S_n=2138.5\)। दिए गए आंकड़ों की संगति भी जांचें। / Adding the two equations gives \(2S_n=4277\), so \(S_n=2138.5\). Also check the consistency of the given data.
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Find the sum of natural numbers from (126) to (175).
A. (7525)
योग \(S_{175}-S_{125}=15400-7875=7525\) है। \(S_b-S_{a-1}\) विधि उपयोग करें। / The sum is \(S_{175}-S_{125}=15400-7875=7525\). Use the \(S_b-S_{a-1}\) method.
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If \(S_n=7875\), what is the value of \(S_{n-7}\)?
C. (7503)
\(7875=S_{125}\), इसलिए \(S_{118}=\frac{118\cdot119}{2}=7021\) नहीं है। सही गणना \(S_{118}=7021\) देती है। / \(7875=S_{125}\), so \(S_{118}=\frac{118\cdot119}{2}=7021\). The listed options should be checked carefully.
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If \(S_{n+2}-S_{n-2}=402\), what will be the value of (n)?
B. (100)
(S_{n+2}-S_{n-2}=(n-1)+n+(n+1)+(n+2)=4n+2)। (4n+2=402) से (n=100)। / (S_{n+2}-S_{n-2}=(n-1)+n+(n+1)+(n+2)=4n+2). From (4n+2=402), (n=100).
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What is (20%) of the sum of the first (150) natural numbers?
C. (2265)
\(S_{150}=11325\) और \(20%=\frac{1}{5}\), इसलिए मान (2265) है। प्रतिशत को भिन्न में बदलें। / \(S_{150}=11325\) and \(20%=\frac{1}{5}\), so the value is (2265). Convert percentage into a fraction.
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If \(S_n=12246\), what is the value of (n)?
C. (156)
\(\frac{156\cdot157}{2}=12246\), इसलिए (n=156)। सम संख्या को आधा करके गुणा करें। / \(\frac{156\cdot157}{2}=12246\), so (n=156). Halve the even number and multiply.
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If \(S_n-S_{70}=5250\), what is the value of (n)?
B. (125)
\(S_{125}-S_{70}=7875-2485=5390\) नहीं है, इसलिए दिए विकल्पों में (5250) का सही मिलान नहीं है। पहले अंतर की जांच करें। / \(S_{125}-S_{70}=7875-2485=5390\), so none of the listed options matches (5250). Check the difference first.
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What is the difference between the sum from (1) to (200) and the sum from (1) to (150)?
B. (8775)
\(S_{200}-S_{150}=20100-11325=8775\)। यह (151) से (200) तक का योग है। / \(S_{200}-S_{150}=20100-11325=8775\). This is the sum from (151) to (200).
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If \(S_n=20100\), what will be the value of \(S_n-S_{n-8}\)?
A. (1572)
\(20100=S_{200}\), इसलिए अंतर \(193+194+\cdots+200=1572\) है। अंतिम (8) पदों का योग लें। / \(20100=S_{200}\), so the difference is \(193+194+\cdots+200=1572\). Take the sum of the last (8) terms.
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If \(S_{3n}=1770\), what is the value of (n)?
B. (19)
\(1770=S_{59}\), इसलिए (3n=59) होगा जो पूर्णांक (n) नहीं देता। सही निष्कर्ष है कि कोई पूर्णांक (n) नहीं है। / \(1770=S_{59}\), so (3n=59), which does not give an integer (n). The correct conclusion is no integer (n).
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What is the sum of natural numbers from (50) to (150)?
A. (10100)
योग \(S_{150}-S_{49}=11325-1225=10100\) है। निचली सीमा से एक कम तक का योग घटाएं। / The sum is \(S_{150}-S_{49}=11325-1225=10100\). Subtract the sum up to one less than the lower limit.
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If \(S_n-S_{n-3}=348\), what is the value of \(S_n\)?
A. (6670)
\(S_n-S_{n-3}=3n-3\), इसलिए (3n-3=348) से (n=117)। अतः \(S_{117}=6903\) होना चाहिए। / \(S_n-S_{n-3}=3n-3\), so (3n-3=348) gives (n=117). Hence \(S_{117}=6903\).
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What is the ratio of the sums of the first (96) and first (72) natural numbers?
B. (194:109)
\(S_{96}:S_{72}=4656:2628=194:109\)। अनुपात को सरलतम रूप में लिखें। / \(S_{96}:S_{72}=4656:2628=194:109\). Write the ratio in simplest form.
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If \(S_n=15400\), what is the value of \(S_{n-10}\)?
C. (13545)
\(15400=S_{175}\), इसलिए \(S_{165}=\frac{165\cdot166}{2}=13695\) है। दिए विकल्पों में सही मान नहीं है। / \(15400=S_{175}\), so \(S_{165}=\frac{165\cdot166}{2}=13695\). The listed options do not contain the correct value.
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If \(S_n+S_{n-2}=10000\) and \(S_n-S_{n-2}=199\), what will \(S_n\) be?
B. (5099.5)
दोनों समीकरण जोड़ने पर \(2S_n=10199\), इसलिए \(S_n=5099.5\)। ऐसे मानों से प्रश्न की संगति भी जांचें। / Adding the two equations gives \(2S_n=10199\), so \(S_n=5099.5\). Such values also help check the consistency of the question.
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How much is the sum of natural numbers from (201) to (250)?
A. (11275)
योग \(S_{250}-S_{200}=31375-20100=11275\) है। बड़े अंतराल में घटाव विधि तेज होती है। / The sum is \(S_{250}-S_{200}=31375-20100=11275\). The subtraction method is quick for large intervals.
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If \(S_{n+1}-S_{n-4}=595\), what will be the value of (n)?
B. (118)
(S_{n+1}-S_{n-4}=(n-3)+(n-2)+(n-1)+n+(n+1)=5n-5)। (5n-5=595) से (n=120)। / (S_{n+1}-S_{n-4}=(n-3)+(n-2)+(n-1)+n+(n+1)=5n-5). From (5n-5=595), (n=120).
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What is the difference between the sums of the first (180) and first (90) natural numbers?
A. (12195)
\(S_{180}-S_{90}=16290-4095=12195\)। यह (91) से (180) तक का योग है। / \(S_{180}-S_{90}=16290-4095=12195\). This is the sum from (91) to (180).
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If \(S_n=8911\), what is the value of (n)?
B. (133)
\(\frac{133\cdot134}{2}=8911\), इसलिए (n=133)। विकल्पों को सूत्र में रखकर सत्यापित करें। / \(\frac{133\cdot134}{2}=8911\), so (n=133). Verify the options by substituting them into the formula.
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If \(S_n=7140\), what is the value of \(S_{n+5}-S_n\)?
C. (605)
\(7140=S_{119}\), इसलिए अंतर (120+121+122+123+124=610) है। सही विकल्प (D) होना चाहिए। / \(7140=S_{119}\), so the difference is (120+121+122+123+124=610). The correct option should be (D).
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What will be the sum of natural numbers from (81) to (144)?
A. (7200)
योग \(S_{144}-S_{80}=10440-3240=7200\) है। (81) से शुरू हो तो (80) तक का योग घटाएं। / The sum is \(S_{144}-S_{80}=10440-3240=7200\). If it starts from (81), subtract the sum up to (80).
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If \(S_n-S_{100}=15150\), what is the value of (n)?
C. (200)
\(S_{200}-S_{100}=20100-5050=15050\), इसलिए दिए विकल्पों में (15150) का सही मिलान नहीं है। अंतर हमेशा जांचें। / \(S_{200}-S_{100}=20100-5050=15050\), so none of the listed options matches (15150). Always check the difference.
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A. (2054)
\(S_{78}-S_{52}+S_{26}=3081-1378+351=2054\)। हर आंशिक योग अलग लिखें। / \(S_{78}-S_{52}+S_{26}=3081-1378+351=2054\). Write each partial sum separately.
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If \(S_n=1225\), what will be the value of \(S_{2n}\)?
C. (4950)
\(1225=S_{49}\), इसलिए (2n=98) और \(S_{98}=\frac{98\cdot99}{2}=4851\)। सही विकल्प (B) होना चाहिए। / \(1225=S_{49}\), so (2n=98) and \(S_{98}=\frac{98\cdot99}{2}=4851\). The correct option should be (B).
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If \(S_{n+4}-S_n=490\), what is the value of (n)?
A. (120)
(S_{n+4}-S_n=(n+1)+(n+2)+(n+3)+(n+4)=4n+10)। (4n+10=490) से (n=120)। / (S_{n+4}-S_n=(n+1)+(n+2)+(n+3)+(n+4)=4n+10). From (4n+10=490), (n=120).
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C. (7525)
परिणाम \(S_{175}-S_{125}=15400-7875=7525\) है। यह (126) से (175) तक का योग है। / The result is \(S_{175}-S_{125}=15400-7875=7525\). This is the sum from (126) to (175).
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What is the ratio of the sums of the first (120) and first (80) natural numbers?
B. (121:81)
\(S_{120}:S_{80}=7260:3240=121:54\), इसलिए दिए विकल्पों में सही सरल अनुपात नहीं है। अनुपात को पूरा सरल करें। / \(S_{120}:S_{80}=7260:3240=121:54\), so the listed options do not contain the correct simplified ratio. Simplify fully.
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If \(S_n=11325\), what will be \(S_n-S_{n-6}\)?
A. (885)
\(11325=S_{150}\), इसलिए अंतर (145+146+147+148+149+150=885) है। अंतिम छह पदों को जोड़ें। / \(11325=S_{150}\), so the difference is (145+146+147+148+149+150=885). Add the last six terms.
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If \(1+2+3+\cdots+n=8256\), what is the value of (n)?
B. (128)
\(\frac{128\cdot129}{2}=8256\), इसलिए (n=128)। उत्तर को सूत्र से जांचना जरूरी है। / \(\frac{128\cdot129}{2}=8256\), so (n=128). It is important to check the answer with the formula.
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What is the sum of natural numbers from (63) to (150)?
A. (9369)
योग \(S_{150}-S_{62}=11325-1953=9372\) है। दिए विकल्पों में सही मान नहीं है। / The sum is \(S_{150}-S_{62}=11325-1953=9372\). The listed options do not contain the correct value.
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If \(S_n-S_{n-3}=420\), what is (n)?
C. (141)
\(S_n-S_{n-3}=3n-3\)। (3n-3=420) से (n=141)। / \(S_n-S_{n-3}=3n-3\). From (3n-3=420), (n=141).
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What is half of the sum of the first (144) natural numbers?
B. (5220)
\(S_{144}=\frac{144\cdot145}{2}=10440\), इसलिए आधा (5220) है। पहले पूर्ण योग निकालें। / \(S_{144}=\frac{144\cdot145}{2}=10440\), so half is (5220). Find the full sum first.
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If \(S_n=6903\), what is the value of \(S_{n+5}-S_n\)?
B. (600)
\(6903=S_{117}\), इसलिए अंतर (118+119+120+121+122=600) है। अगले पांच पदों का योग लें। / \(6903=S_{117}\), so the difference is (118+119+120+121+122=600). Take the sum of the next five terms.
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Find the sum of natural numbers from (151) to (225).
B. (14100)
योग \(S_{225}-S_{150}=25425-11325=14100\) है। \(S_b-S_{a-1}\) विधि प्रयोग करें। / The sum is \(S_{225}-S_{150}=25425-11325=14100\). Use the \(S_b-S_{a-1}\) method.
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If \(S_n+S_{n-1}=1260\) and \(S_n-S_{n-1}=36\), which is the value of \(S_n\)?
B. (648)
दोनों समीकरण जोड़ने पर \(2S_n=1296\), इसलिए \(S_n=648\)। समीकरणों को जोड़कर \(S_n\) जल्दी मिलता है। / Adding the two equations gives \(2S_n=1296\), so \(S_n=648\). Adding equations gives \(S_n\) quickly.
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If \(S_{n+1}-S_{n-2}=330\), what will be the value of (n)?
C. (110)
(S_{n+1}-S_{n-2}=(n-1)+n+(n+1)=3n)। (3n=330) से (n=110)। / (S_{n+1}-S_{n-2}=(n-1)+n+(n+1)=3n). From (3n=330), (n=110).
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B. (180)
\(S_{180}-S_{179}=180\) होता है। क्रमिक योगों का अंतर अगला पद होता है। / \(S_{180}-S_{179}=180\). The difference of consecutive sums is the next term.
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What is the difference between the sum from (1) to (250) and the sum from (1) to (200)?
B. (11275)
\(S_{250}-S_{200}=31375-20100=11275\)। यह (201) से (250) तक का योग है। / \(S_{250}-S_{200}=31375-20100=11275\). This is the sum from (201) to (250).
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If \(S_n=7626\), what is the value of \(S_{n+2}\)?
A. (7875)
\(7626=S_{123}\), इसलिए \(S_{125}=\frac{125\cdot126}{2}=7875\)। पहले (n) पहचानें फिर (2) आगे बढ़ें। / \(7626=S_{123}\), so \(S_{125}=\frac{125\cdot126}{2}=7875\). First identify (n) and then move (2) ahead.
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If \(S_n-S_{n-1}=150\), what will be \(S_{n-1}\)?
A. (11175)
\(S_n-S_{n-1}=n\), इसलिए (n=150) और \(S_{149}=\frac{149\cdot150}{2}=11175\)। पिछले योग के लिए (n-1) लें। / \(S_n-S_{n-1}=n\), so (n=150) and \(S_{149}=\frac{149\cdot150}{2}=11175\). For the previous sum, take (n-1).
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What is (40%) of the sum of the first (125) natural numbers?
A. (3150)
\(S_{125}=7875\) और \(40%=\frac{2}{5}\), इसलिए मान (3150) है। प्रतिशत को भिन्न में बदलें। / \(S_{125}=7875\) and \(40%=\frac{2}{5}\), so the value is (3150). Convert percentage into a fraction.
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If \(S_n=25425\), what will be the value of (n)?
C. (225)
\(\frac{225\cdot226}{2}=25425\), इसलिए (n=225)। सूत्र में विकल्प रखकर उत्तर जांचें। / \(\frac{225\cdot226}{2}=25425\), so (n=225). Check the answer by substituting options in the formula.
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If \(S_n-S_{n-6}=861\), what will be the value of (n)?
D. (146)
(S_n-S_{n-6}=(n-5)+(n-4)+(n-3)+(n-2)+(n-1)+n=6n-15)। (6n-15=861) से (n=146)। / (S_n-S_{n-6}=(n-5)+(n-4)+(n-3)+(n-2)+(n-1)+n=6n-15). From (6n-15=861), (n=146).
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If (S_n=\frac{n(n+1)}{2}) and \(S_n=7503\), what is the value of (n)?
C. (122)
\(\frac{122\cdot123}{2}=7503\), इसलिए (n=122)। बड़े योग में विकल्पों को सूत्र में जांचें। / \(\frac{122\cdot123}{2}=7503\), so (n=122). For a large sum, check options in the formula.
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What is the sum of natural numbers from (56) to (120)?
B. (5720)
योग \(S_{120}-S_{55}=7260-1540=5720\) है। निचली सीमा से एक कम तक का योग घटाएं। / The sum is \(S_{120}-S_{55}=7260-1540=5720\). Subtract the sum up to one less than the lower limit.
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If \(S_n-S_{n-6}=813\), what will be the value of (n)?
D. (138)
\(S_n-S_{n-6}=6n-15\) होता है। (6n-15=813) से (n=138)। / \(S_n-S_{n-6}=6n-15\). From (6n-15=813), (n=138).
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If \(S_{n+4}-S_{n-1}=485\), what is the value of (n)?
A. (95)
यह अंतर (n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10) है। (5n+10=485) से (n=95)। / This difference is (n+(n+1)+(n+2)+(n+3)+(n+4)=5n+10). From (5n+10=485), (n=95).
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If \(S_n=9870\), what is the value of (n)?
B. (140)
\(\frac{140\cdot141}{2}=9870\), इसलिए (n=140)। सम संख्या को आधा करके गुणा करें। / \(\frac{140\cdot141}{2}=9870\), so (n=140). Halve the even number and multiply.
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If \(S_n=2701\), what will be the value of \(S_{n+7}-S_n\)?
D. (539)
\(2701=S_{73}\), इसलिए अंतर \(74+75+\cdots+80=539\) है। अगले सात पदों का योग लें। / \(2701=S_{73}\), so the difference is \(74+75+\cdots+80=539\). Take the sum of the next seven terms.
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What is the ratio of the sums of the first (84) and first (56) natural numbers?
A. (85:38)
\(S_{84}:S_{56}=3570:1596=85:38\)। अनुपात को सरलतम रूप में लिखें। / \(S_{84}:S_{56}=3570:1596=85:38\). Write the ratio in simplest form.
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If \(S_n+S_{n-1}=5625\) and \(S_n-S_{n-1}=75\), what is the value of \(S_n\)?
B. (2850)
दोनों समीकरण जोड़ने पर \(2S_n=5700\), इसलिए \(S_n=2850\)। समीकरणों को जोड़ना तेज तरीका है। / Adding the two equations gives \(2S_n=5700\), so \(S_n=2850\). Adding equations is a quick method.
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How much is the sum of natural numbers from (101) to (160)?
C. (7830)
योग \(S_{160}-S_{100}=12880-5050=7830\) है। बड़े अंतराल में घटाव विधि उपयोग करें। / The sum is \(S_{160}-S_{100}=12880-5050=7830\). Use the subtraction method for large intervals.
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If \(S_n=16290\), what is the value of \(S_{n-8}\)?
D. (14878)
\(16290=S_{180}\), इसलिए \(S_{172}=\frac{172\cdot173}{2}=14878\)। पहले (n) पहचानें फिर (8) कम करें। / \(16290=S_{180}\), so \(S_{172}=\frac{172\cdot173}{2}=14878\). Identify (n) first and then reduce it by (8).
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If \(S_{n+2}-S_{n-3}=625\), what is the value of (n)?
A. (125)
यह अंतर ((n-2)+(n-1)+n+(n+1)+(n+2)=5n) है। इसलिए (5n=625) से (n=125)। / This difference is ((n-2)+(n-1)+n+(n+1)+(n+2)=5n). So (5n=625) gives (n=125).
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What is (30%) of the sum of the first (200) natural numbers?
C. (6030)
\(S_{200}=20100\) और \(30%=\frac{3}{10}\), इसलिए मान (6030) है। प्रतिशत को भिन्न में बदलें। / \(S_{200}=20100\) and \(30%=\frac{3}{10}\), so the value is (6030). Convert the percentage into a fraction.
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If \(S_n=11175\), what is the value of (n)?
A. (149)
\(\frac{149\cdot150}{2}=11175\), इसलिए (n=149)। विकल्पों को सूत्र में रखकर जांचें। / \(\frac{149\cdot150}{2}=11175\), so (n=149). Check the options in the formula.
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If \(S_n-S_{90}=5775\), what will be the value of (n)?
D. (140)
\(S_{140}-S_{90}=9870-4095=5775\), इसलिए (n=140)। दिए अंतर को \(S_n-S_k\) से मिलाएं। / \(S_{140}-S_{90}=9870-4095=5775\), so (n=140). Match the given difference with \(S_n-S_k\).
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What is the sum of natural numbers from (151) to (210)?
B. (10830)
योग \(S_{210}-S_{150}=22155-11325=10830\) है। \(S_b-S_{a-1}\) विधि का उपयोग करें। / The sum is \(S_{210}-S_{150}=22155-11325=10830\). Use the \(S_b-S_{a-1}\) method.
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If \(S_{2n}=4095\), what is the value of (n)?
C. (45)
\(4095=S_{90}\), इसलिए (2n=90) और (n=45)। पहले सूचकांक पहचानें। / \(4095=S_{90}\), so (2n=90) and (n=45). Identify the index first.
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If \(S_n-S_{n-4}=454\), what is the value of (n)?
D. (115)
\(S_n-S_{n-4}=4n-6\) होता है। (4n-6=454) से (n=115)। / \(S_n-S_{n-4}=4n-6\). From (4n-6=454), (n=115).
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What is the difference between the sums of the first (225) and first (175) natural numbers?
A. (10025)
\(S_{225}-S_{175}=25425-15400=10025\)। यह (176) से (225) तक का योग है। / \(S_{225}-S_{175}=25425-15400=10025\). This is the sum from (176) to (225).
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What is the ratio of the sums of the first (150) and first (100) natural numbers?
B. (453:202)
\(S_{150}:S_{100}=11325:5050=453:202\)। अनुपात को पूरा सरल करें। / \(S_{150}:S_{100}=11325:5050=453:202\). Simplify the ratio fully.
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If \(S_n=15400\), what will be the value of \(S_n-S_{n-10}\)?
C. (1705)
\(15400=S_{175}\), इसलिए अंतिम दस पदों का योग \(166+167+\cdots+175=1705\) है। अंतिम पदों को सीधे जोड़ें। / \(15400=S_{175}\), so the sum of the last ten terms is \(166+167+\cdots+175=1705\). Add the last terms directly.
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If \(S_{n+1}-S_{n-4}=695\), what will be the value of (n)?
D. (140)
यह अंतर ((n-3)+(n-2)+(n-1)+n+(n+1)=5n-5) है। (5n-5=695) से (n=140)। / This difference is ((n-3)+(n-2)+(n-1)+n+(n+1)=5n-5). From (5n-5=695), (n=140).
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What is the sum of natural numbers from (75) to (125)?
A. (5100)
योग \(S_{125}-S_{74}=7875-2775=5100\) है। निचली सीमा से पहले का योग घटाएं। / The sum is \(S_{125}-S_{74}=7875-2775=5100\). Subtract the sum before the lower limit.
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What is the value of \(S_{120}-S_{80}+S_{40}\)?
B. (4840)
\(S_{120}-S_{80}+S_{40}=7260-3240+820=4840\)। हर आंशिक योग अलग लिखें। / \(S_{120}-S_{80}+S_{40}=7260-3240+820=4840\). Write each partial sum separately.
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If \(S_n=10440\), what is the value of (n)?
C. (144)
\(\frac{144\cdot145}{2}=10440\), इसलिए (n=144)। सूत्र में विकल्प रखकर उत्तर जांचें। / \(\frac{144\cdot145}{2}=10440\), so (n=144). Check the answer by substituting options in the formula.
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If \(S_n-S_{n-2}=257\), what is the value of \(S_n\)?
D. (8385)
\(S_n-S_{n-2}=2n-1\) है। (2n-1=257) से (n=129), अतः \(S_{129}=8385\)। / \(S_n-S_{n-2}=2n-1\). From (2n-1=257), (n=129), so \(S_{129}=8385\).
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Find the sum of natural numbers from (181) to (240).
A. (12630)
योग \(S_{240}-S_{180}=28920-16290=12630\) है। बड़े अंतराल में घटाव विधि तेज होती है। / The sum is \(S_{240}-S_{180}=28920-16290=12630\). The subtraction method is quick for large intervals.
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What is the ratio of the sums of the first (128) and first (64) natural numbers?
C. (258:65)
\(S_{128}:S_{64}=8256:2080=258:65\)। अनुपात को सरलतम रूप में लिखें। / \(S_{128}:S_{64}=8256:2080=258:65\). Write the ratio in simplest form.
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If \(S_n=8256\), what will be \(S_{n+5}-S_n\)?
B. (655)
\(8256=S_{128}\), इसलिए अंतर (129+130+131+132+133=655) है। अगले पांच पदों को जोड़ें। / \(8256=S_{128}\), so the difference is (129+130+131+132+133=655). Add the next five terms.
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If \(S_n+S_{n-1}=12544\) and \(S_n-S_{n-1}=112\), what is the value of \(S_n\)?
C. (6328)
दोनों समीकरण जोड़ने पर \(2S_n=12656\), इसलिए \(S_n=6328\)। समीकरणों को जोड़ना सबसे तेज है। / Adding the two equations gives \(2S_n=12656\), so \(S_n=6328\). Adding the equations is the fastest way.
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If \(S_n=1830\), what will be the value of \(S_{3n}\)?
D. (16290)
\(1830=S_{60}\), इसलिए (3n=180) और \(S_{180}=16290\)। पहले (n) पहचानें। / \(1830=S_{60}\), so (3n=180) and \(S_{180}=16290\). Identify (n) first.
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If \(S_{n+3}-S_{n-2}=760\), what is the value of (n)?
A. (151)
यह अंतर ((n-1)+n+(n+1)+(n+2)+(n+3)=5n+5) है। (5n+5=760) से (n=151)। / This difference is ((n-1)+n+(n+1)+(n+2)+(n+3)=5n+5). From (5n+5=760), (n=151).
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What is the sum of natural numbers from (90) to (150)?
B. (7320)
योग \(S_{150}-S_{89}=11325-4005=7320\) है। (90) से शुरू हो तो (89) तक का योग घटाएं। / The sum is \(S_{150}-S_{89}=11325-4005=7320\). If it starts from (90), subtract the sum up to (89).
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What is (40%) of the sum of the first (175) natural numbers?
C. (6160)
\(S_{175}=15400\) और \(40%=\frac{2}{5}\), इसलिए मान (6160) है। प्रतिशत को भिन्न में बदलें। / \(S_{175}=15400\) and \(40%=\frac{2}{5}\), so the value is (6160). Convert the percentage into a fraction.
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If \(S_n=19900\), what is the value of (n)?
D. (199)
\(\frac{199\cdot200}{2}=19900\), इसलिए (n=199)। सम संख्या को आधा करके गुणा करें। / \(\frac{199\cdot200}{2}=19900\), so (n=199). Halve the even number and multiply.
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If \(S_n-S_{n-7}=1330\), what will be the value of (n)?
A. (193)
\(S_n-S_{n-7}=7n-21\) होता है। (7n-21=1330) से (n=193)। / \(S_n-S_{n-7}=7n-21\). From (7n-21=1330), (n=193).
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How much is the sum of natural numbers from (211) to (260)?
B. (11775)
योग \(S_{260}-S_{210}=33930-22155=11775\) है। बड़े अंतराल में \(S_b-S_{a-1}\) प्रयोग करें। / The sum is \(S_{260}-S_{210}=33930-22155=11775\). For large intervals, use \(S_b-S_{a-1}\).
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What is the ratio of the sums of the first (100) and first (50) natural numbers?
C. (202:51)
\(S_{100}:S_{50}=5050:1275=202:51\)। अनुपात को पूरी तरह सरल करें। / \(S_{100}:S_{50}=5050:1275=202:51\). Simplify the ratio fully.
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If \(S_n=11175\), what will be \(S_n-S_{n-4}\)?
D. (590)
\(11175=S_{149}\), इसलिए अंतर (146+147+148+149=590) है। अंतिम चार पद जोड़ें। / \(11175=S_{149}\), so the difference is (146+147+148+149=590). Add the last four terms.
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If (n=50), what is the value of \(S_{2n}-S_n\)?
A. (3775)
\(S_{2n}-S_n=S_{100}-S_{50}=5050-1275=3775\)। पहले (2n) निकालें। / \(S_{2n}-S_n=S_{100}-S_{50}=5050-1275=3775\). Find (2n) first.
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Find the sum of natural numbers from (301) to (350).
B. (16275)
योग \(S_{350}-S_{300}=61425-45150=16275\) है। (301) से शुरू हो तो (300) तक का योग घटाएं। / The sum is \(S_{350}-S_{300}=61425-45150=16275\). If it starts from (301), subtract the sum up to (300).
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If \(S_n-S_{n-2}=199\), what will be the value of \(S_n\)?
C. (5050)
\(S_n-S_{n-2}=2n-1\) है। (2n-1=199) से (n=100), इसलिए \(S_n=5050\)। / \(S_n-S_{n-2}=2n-1\). From (2n-1=199), (n=100), so \(S_n=5050\).
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If \(S_n=13041\), what is the value of (n)?
D. (161)
\(\frac{161\cdot162}{2}=13041\), इसलिए (n=161)। विकल्पों को सूत्र में रखकर जांचें। / \(\frac{161\cdot162}{2}=13041\), so (n=161). Check the options in the formula.
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If \(S_{n+5}-S_{n-5}=1205\), what will be the value of (n)?
A. (120)
यह अंतर ((n-4)+(n-3)+\cdots+(n+5)=10n+5) है। (10n+5=1205) से (n=120)। / This difference is ((n-4)+(n-3)+\cdots+(n+5)=10n+5). From (10n+5=1205), (n=120).
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What is the difference between the sums of the first (180) and first (120) natural numbers?
B. (9030)
\(S_{180}-S_{120}=16290-7260=9030\)। यह (121) से (180) तक का योग है। / \(S_{180}-S_{120}=16290-7260=9030\). This is the sum from (121) to (180).
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If \(S_n=2145\), what will be the value of \(S_{2n}-S_n\)?
C. (6370)
\(2145=S_{65}\), इसलिए \(S_{2n}-S_n=S_{130}-S_{65}=8515-2145=6370\)। पहले (n) पहचानें। / \(2145=S_{65}\), so \(S_{2n}-S_n=S_{130}-S_{65}=8515-2145=6370\). Identify (n) first.
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What is the sum of natural numbers from (401) to (450)?
A. (21275)
योग \(S_{450}-S_{400}=101475-80200=21275\) है। बड़े अंतराल में घटाव विधि उपयोग करें। / The sum is \(S_{450}-S_{400}=101475-80200=21275\). Use the subtraction method for large intervals.
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If \(S_n-S_{n-1}=225\), what will be the value of \(S_{n+1}\)?
B. (25651)
\(S_n-S_{n-1}=n\), इसलिए (n=225) और \(S_{226}=25651\)। अगले योग के लिए (n+1) लें। / \(S_n-S_{n-1}=n\), so (n=225) and \(S_{226}=25651\). For the next sum, take (n+1).
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If \(S_n=31375\), what is the value of \(S_{n-12}\)?
C. (28441)
\(31375=S_{250}\), इसलिए \(S_{238}=\frac{238\cdot239}{2}=28441\)। पहले (n) पहचानकर (12) कम करें। / \(31375=S_{250}\), so \(S_{238}=\frac{238\cdot239}{2}=28441\). Identify (n) first and reduce it by (12).
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If \(S_n-S_{n-8}=1228\), what will be the value of (n)?
C. (157)
(S_n-S_{n-8}=(n-7)+(n-6)+\cdots+n=8n-28) होता है। (8n-28=1228) से (n=157)। / (S_n-S_{n-8}=(n-7)+(n-6)+\cdots+n=8n-28). From (8n-28=1228), (n=157).
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Find the sum of natural numbers from (251) to (300).
B. (13775)
योग \(S_{300}-S_{250}=45150-31375=13775\) है। \(S_b-S_{a-1}\) विधि बड़े अंतराल में तेज रहती है। / The sum is \(S_{300}-S_{250}=45150-31375=13775\). The \(S_b-S_{a-1}\) method is fast for large intervals.
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