Suppose that the number of terms in an A.P. is $2k$, $k\in N$. 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 :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Even-position terms minus odd-position terms: $kd=55-40=15$. Last minus first: $(2k-1)d=27$. Dividing, $\frac{2k-1}{k}=\frac95\Rightarrow k=5$. Answer (2).

Q. Suppose that the number of terms in an A.P. is $2k$, $k\in N$. 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 :

Explanation

Reviewed by Dr. Abubacker

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