Question

If the ratio of the sum of $n$ terms of two A.P.'s be $( 7 n + 1 ) : ( 4 n + 27 )$, then the ratio of their terms will be.

• A. $2 : 3$
• B. $3 : 4$
• C. $4 : 3$
• D. $5 : 6$

Answer 1

Answer: (C)

$4 : 3$

Answer 2

Let $S _ { n }$ and be the sums of n terms of two A.P.'s and $T _ { 11 }$ and be the respective terms, then

$\frac { S _ { n } } { S _ { n } ^ { \prime } } = \frac { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \frac { n } { 2 } \left[ 2 a ^ { \prime } + ( n - 1 ) d ^ { \prime } \right] } = \frac { 7 n + 1 } { 4 n + 27 }$

$\Rightarrow$ $\frac { a + \frac { ( n - 1 ) } { 2 } d } { a ^ { \prime } + \frac { ( n - 1 ) } { 2 } d ^ { \prime } } = \frac { 7 n + 1 } { 4 n + 27 }$

Now put $n = 21$ ,

we get $\frac { a + 10 d } { a ^ { \prime } + 10 d ^ { \prime } } = \frac { T _ { 11 } } { T _ { 11 } ^ { \prime } } = \frac { 148 } { 111 } = \frac { 4 } { 3 }$ .

Note : If ratio of sum of $n$ terms of two A.P.'s are given in terms of $n$ and ratio of their $p ^ { t h }$ terms are to be found then put $n = 2 p - 1$. Here we put $n = 11 \times 2 - 1 = 21$.