Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 181-190 of 600 questions.
What is (4) times the sum of the first (21) natural numbers?
C. (924)
\(S_{21}=231\) है और \(4\times231=924\)। पहले योग निकालकर फिर गुणा करें। / \(S_{21}=231\), and \(4\times231=924\). First find the sum and then multiply.
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If \(S_n=1431\), what will be the value of (n)?
C. (53)
\(S_{53}=\frac{53\times54}{2}=1431\) इसलिए (n=53) है। \(2S_n\) को क्रमागत गुणनफल से मिलाएं। / \(S_{53}=\frac{53\times54}{2}=1431\), so (n=53). Match \(2S_n\) with a consecutive product.
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What is the difference between the sum of the first (39) natural numbers and (39)?
C. (741)
\(S_{39}=780\) और (780-39=741) है। अंतिम पद घटाने पर \(S_{38}\) मिलता है। / \(S_{39}=780\), and (780-39=741). Subtracting the last term gives \(S_{38}\).
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What is the sum of \(23+24+25+\cdots+34\)?
A. (342)
योग \(S_{34}-S_{22}=595-253=342\) है। आरंभ से पहले वाले पदों का योग घटाएं। / The sum is \(S_{34}-S_{22}=595-253=342\). Subtract the sum of terms before the start.
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For which (n) will \(S_n-S_{n-1}=62\)?
C. (62)
\(S_n-S_{n-1}=n\) होता है इसलिए (n=62)। दो लगातार योगों का अंतर नया पद होता है। / \(S_n-S_{n-1}=n\), so (n=62). The difference of two consecutive sums is the new term.
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Find the sum of the first (75) natural numbers.
A. (2850)
\(S_{75}=\frac{75\times76}{2}=2850\) है। (76) को (38) बनाकर गुणा करें। / \(S_{75}=\frac{75\times76}{2}=2850\). Change (76) to (38) and multiply.
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If \(S_{18}=171\), what will be the value of \(S_{22}\)?
B. (253)
\(S_{22}=S_{18}+19+20+21+22=253\) है। आगे के सभी नए पद जोड़ना जरूरी है। / \(S_{22}=S_{18}+19+20+21+22=253\). Adding all new terms ahead is necessary.
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What is obtained by adding (46), (47), and (48) to the sum of the first (45) natural numbers?
C. \(S_{48}\)
\(S_{45}+46+47+48=S_{48}\) होता है। लगातार पद जोड़ने पर नया अंतिम पद सूचकांक बनता है। / \(S_{45}+46+47+48=S_{48}\). When consecutive terms are added, the new last term becomes the index.
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What is the ratio of \(S_{30}\) to \(S_{15}\)?
A. (31:8)
\(S_{30}:S_{15}=465:120=31:8\) है। अनुपात को समान गुणक से भाग देकर सरल करें। / \(S_{30}:S_{15}=465:120=31:8\). Simplify the ratio by dividing by a common factor.
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B. (406)
कुल फूल \(S_{28}=\frac{28\times29}{2}=406\) हैं। त्रिभुजाकार पंक्तियां \(S_n\) का संकेत देती हैं। / Total flowers are \(S_{28}=\frac{28\times29}{2}=406\). Triangular rows indicate \(S_n\).
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