Question

If first, fifth and last terms of an A.P. is l, m, p respectively and sum of the A.P. is $\frac{(\mathcal{l} + p)(4p + m - s\mathcal{l})}{k(m\mathcal{- l})}$then k is –

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (A)

2

Answer 2

Let common difference = d & number of terms = n

\ T₅: m = l + (5 –1) d ̃ d = (m –l) /4

\ T_n: p = l + $\frac{(n - 1)(m\mathcal{- l})}{4}$

̃ n = $\left( \frac{4p + m - 5\mathcal{l}}{m\mathcal{- l}} \right)$

\ Sum of n terms of A.P. = $\frac{n}{2}$ [First term + last term]

= $\left\lbrack \frac{4p + m - 5\mathcal{l}}{m\mathcal{- l}} \right\rbrack$. $\frac{1}{2}$ [l + p] …(i)

Comparing equation (i) with the given summation, we get

k = 2.