Verified Solution • 2025–26 Session moderate

QJEEM25Maa19b: JEE Main 2025 Mathematics - Class 11 JEEM25Maa19b Solution: Suppose that the number of terms in an A.P. is . If the sum of…

Suppose that the number of terms in an A.P. is 2k,k∈N2 k, k \in N . If the sum of all odd terms of the A.P. is 40 , the sum of all even terms is 55 and the last term of the A.P. exceeds the first term by 27 , then k is equal to:

Multiple Choice Options

  • A 8
  • B 6
  • C 4
  • D 5

Step-by-Step Verified Solution

amp;a1,a2,a3,……,a2k→A.P.amp;∑r=1ka2r−1=40,∑r=1ka2r=55,a2k−a1=27amp;k2[2a1+(k−1)2 d]=40,k2[2a2+(k−1)2 d]=55,amp; d=272k−1amp;a1=40k−(k−1)d=55k−kdamp; d=15k⇒272k−1=15k⇒9k=10k−5amp;∴k=5\begin{aligned}& a_1, a_2, a_3, \ldots \ldots, a_{2 k} \quad \rightarrow \mathrm{A.P.}\\& \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}-1}=40, \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}}=55, \mathrm{a}_{2 \mathrm{k}}-\mathrm{a}_1=27 \\& \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_1+(\mathrm{k}-1) 2 \mathrm{~d}\right]=40, \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_2+(\mathrm{k}-1) 2 \mathrm{~d}\right]=55, \\& \mathrm{~d}=\frac{27}{2 \mathrm{k}-1} \\& \mathrm{a}_1=\frac{40}{\mathrm{k}}-(\mathrm{k}-1) \mathrm{d}=\frac{55}{\mathrm{k}}-\mathrm{kd} \\& \mathrm{~d}=\frac{15}{\mathrm{k}} \Rightarrow \frac{27}{2 \mathrm{k}-1}=\frac{15}{\mathrm{k}} \Rightarrow 9 \mathrm{k}=10 \mathrm{k}-5 \\& \therefore \mathrm{k}=5\end{aligned}

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