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Mathematics · 2025 · 5 marks
CBSE 2025 · Region 3 · Set 2 · Q32
(a) An AP consists of 'n' terms whose n^th term is 4 and the common difference is 2. If the sum of ' n ' terms of AP is -14, then find ' n '. Also,…
(a)
An AP consists of 'n' terms whose $\displaystyle \mathrm{n}^{\text {th }}$ term is $\displaystyle 4$ and the common difference is $\displaystyle 2$ . If the sum of ' $\displaystyle \mathrm{n}$ ' terms of AP is -$\displaystyle 14$ , then find ' $\displaystyle \mathrm{n}$ '. Also, find the sum of the first $\displaystyle 20$ terms.
OR
(b)
The sum of the first six terms of an arithmetic progression is 42. The ratio of the $\displaystyle 10^{\text {th }}$ term to the $\displaystyle 30^{\text {th }}$ term is $\displaystyle 1: 3$. Calculate the first and the thirteenth terms of the AP.
Marking-scheme solution
Let first term \(\displaystyle =\mathrm{a}\), common difference \(\displaystyle =\mathrm{d}=2\)
ATQ, \(\displaystyle \mathrm{a}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) 2=4\)
\(\displaystyle \mathrm{a}+2 \mathrm{n}=6\)
\(\displaystyle \mathrm{a}=6-2 \mathrm{n}\)
ATQ, \(\displaystyle \mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) 2]=-14\)
\(\displaystyle \mathrm{n}[\mathrm{a}+\mathrm{n}-1]=-14\)
\(\displaystyle \mathrm{n}[6-2 \mathrm{n}+\mathrm{n}-1]=-14\)
\(\displaystyle \mathrm{n}^{2}-5 \mathrm{n}-14=0\)
\(\displaystyle \Rightarrow \mathrm{n}=7\)
and \(\displaystyle \mathrm{a}=-8\)
\(\displaystyle \mathrm{S}_{20}=\frac{20}{2}[2 \times(-8)+19 \times 2]\)
\(\displaystyle =220\)
Let first term \(\displaystyle =a\) and common difference \(\displaystyle =d\)
ATQ, \(\displaystyle \frac{a_{10}}{a_{30}}=\frac{a+9 d}{a+29 d}=\frac{1}{3}\)
\(\displaystyle 3 a+27 d=a+29 d\)
\(\displaystyle a=d\)\(\displaystyle \mathrm{S}_{6}=\frac{6}{2}[2 a+(6-1) a]=42\)
\(\displaystyle a=2\)\(\displaystyle d=2\)\(\displaystyle a_{13}=2+12 \times 2=26\)
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CBSE Class 10 Mathematics past-paper question from the 2025 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.