sequences se related questions ko ek jagah revise karein. Har question me bilingual content, answer feedback aur explanation available hai.
3683 questions tagged with sequences.
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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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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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 \(S_n-S_{n-1}=225\), 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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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 \(S_n=2145\), 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 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 \(S_{n+5}-S_{n-5}=1205\), 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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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 \(S_n-S_{n-2}=199\), 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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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 (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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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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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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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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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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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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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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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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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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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 \(S_n+S_{n-1}=12544\) 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 \(S_n=8256\), 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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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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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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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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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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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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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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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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