Exams Joint Entrance Examination Mathematics Sequences and Series Question 47 Question 47 of 519 Medium MCQ
Suppose that the number of terms in an A.P. is 2 k , k ∈ N 2 k, k \in N 2 k , k ∈ 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:
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\[ a 1 , a 2 , a 3 , … … , a 2 k → A . P . ∑ r = 1 k a 2 r − 1 = 40 , ∑ r = 1 k a 2 r = 55 , a 2 k − a 1 = 27 k 2 [ 2 a 1 + ( k − 1 ) 2 d ] = 40 , k 2 [ 2 a 2 + ( k − 1 ) 2 d ] = 55 , d = 27 2 k − 1 a 1 = 40 k − ( k − 1 ) d = 55 k − k d d = 15 k ⇒ 27 2 k − 1 = 15 k ⇒ 9 k = 10 k − 5 ∴ k = 5 \begin{aligned}
& a_1, a_2, a_3, \ldots \ldots, a_{2 k} \quad \rightarrow \mathrm{A.P.}\\
& \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}-1}=40, \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}}=55, \mathrm{a}_{2 \mathrm{k}}-\mathrm{a}_1=27 \\
& \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_1+(\mathrm{k}-1) 2 \mathrm{~d}\right]=40, \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_2+(\mathrm{k}-1) 2 \mathrm{~d}\right]=55, \\
& \mathrm{~d}=\frac{27}{2 \mathrm{k}-1} \\
& \mathrm{a}_1=\frac{40}{\mathrm{k}}-(\mathrm{k}-1) \mathrm{d}=\frac{55}{\mathrm{k}}-\mathrm{kd} \\
& \mathrm{~d}=\frac{15}{\mathrm{k}} \Rightarrow \frac{27}{2 \mathrm{k}-1}=\frac{15}{\mathrm{k}} \Rightarrow 9 \mathrm{k}=10 \mathrm{k}-5 \\
& \therefore \mathrm{k}=5
\end{aligned} a 1 , a 2 , a 3 , …… , a 2 k → A.P. r = 1 ∑ k a 2 r − 1 = 40 , r = 1 ∑ k a 2 r = 55 , a 2 k − a 1 = 27 2 k [ 2 a 1 + ( k − 1 ) 2 d ] = 40 , 2 k [ 2 a 2 + ( k − 1 ) 2 d ] = 55 , d = 2 k − 1 27 a 1 = k 40 − ( k − 1 ) d = k 55 − kd d = k 15 ⇒ 2 k − 1 27 = k 15 ⇒ 9 k = 10 k − 5 ∴ k = 5 \] Answer the question to unlock