Question

Let the sum of the first $n$ terms of a non-constant $A.P$., $a_1,a_2,a_3,......$be 50n+ $\frac{n(n-7)}{2}A$. where $A$ is a constant. If d is the common difference of this $A.P$., then the ordered pair $(d,a_{50})$ is equal to

• A. $(A,50+46A)$
• B. $(A,50+45A)$
• C. $(50,50+46A)$
• D. $(50,50+45A)$

Answer 1

Answer: (A)

$S_n=50n+\frac{n(n-7)}{2}$A
$T_n=S_n-S_{n-1}$
$=50n+\frac{n(n-7)}{2}A-50(n-1)-\frac{(n-1(n-8)}{2}A$
$=50+\frac{A}{2}[n^2-7n-n^2+9n-8]$
= 50 + A(n - 4)
$d=T_n-T_{n-1}$
= $50+A(n-4)-50-A(n-5)$
= $A$
$T_{50}=50+46A$
$(d.A_{50})=(A,50+46A)$