Level 61 • 50/50 questions • 25 seconds per question.
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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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