Question

If ${(p+1)}^{th}$ term of an A.P.. is twice its ${(q+1)}^{th}$ term, then ${(3p+1)}^{th}$ term

• A. Thrice the ${(p+q)}^{th}$ term
• B. Twice the ${(p+q-1)}^{th}$ term
• C. Twice the ${(p+q+1)}^{th}$ term
• D. Thrice the ${(p+q+1)}^{th}$ term

Answer 1

The correct option is C Twice the ${(p+q+1)}^{th}$ term

Assuming first term of given A.P $a$ and common difference $d$
we know that $n^{th}$ term of an A.P is given by, $t_n$ $=$ $a+(n-1)d$
so $(p+1{)}^{th}$ term, $t_{p+1}$ $=$ $a+pd$
and $(q+1{)}^{th}$ term, $t_{q+1}$ $=$ $a+qd$
given than $t_{p+1}$ $=$ 2.$t_{q+1}$ $\Rightarrow$ $a+pd$ $=$ 2($a+qd$ ) $\to$ (1)
Now,
$(3p+1{)}^{th}$ term, $t_{3p+1}$ $=$ $a+3pd$ $=$ $a+pd+2pd$
using equation (1)
$t_{3p+1}$ $=$ $a+3pd$ $=$ $2(a+qd)+2pd$ $=$ $2[a+(p+q\left)d\right]$ $=$ $2.$$t_{p+q+1}$
so option (c) is correct choice