Question

If the sum of the 10 terms of an A.P. is 4 times tp the sum of its 5 terms, then the ratio of the first term and common difference is
1). 1: 2
2). 2: 1
3). 2: 3
4). 3: 2

Answer 1

Hint: We have given that sum of the first 10 terms of A.P. is 4 times the sum of first 5 terms so, we will find the value of sum of first 10 terms and sum of first 5 terms and then, put them in the equation which is made from the conditions given in the question. Then, by solving this we will get the ratio.

Complete step-by-step solution:
Given: $S_10 = 4 S_5$-----(1)
Formula of finding the sum of terms in A.P. is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
Here, a is the first term and d is the common difference.
Now, we will find the sum of the first 10 terms and sum of the first 5 terms and put them in equation (1).
Sum of first 10 terms:
${S_{10}} = \dfrac{{10}}{2}\left[ {2a + \left( {10 - 1} \right)d} \right]$
$\Rightarrow {S_{10}} = 5\left[ {2a + 9d} \right]$
$\Rightarrow {S_{10}} = 10a + 45d$
Sum of first 5 terms:
$\Rightarrow {S_5} = \dfrac{5}{2}\left[ {2a + \left( {5 - 1} \right)d} \right]$
$\Rightarrow {S_5} = \dfrac{5}{2}\left[ {2a + 4d} \right]$
$\Rightarrow {S_5} = 5a + 10d$
Putting these values in equation (1).
$10a + 45d = 4\left( {5a + 10d} \right)$
$\Rightarrow 10a + 45d = 20a + 40d$
$\Rightarrow 5d = 10a$
$\Rightarrow \dfrac{a}{d} = \dfrac{5}{{10}}$
$\Rightarrow \dfrac{a}{d} = \dfrac{1}{2}$
$\Rightarrow a:d = 1:2$
The ratio of first term and common difference is 1: 2.
So, option (1) is the correct answer.

Note: Whenever we face such types of problems just simply by using the condition provided in question and basic formulas of arithmetic progression (A.P). This will help to reach the solution easily.
Arithmetic progression (A.P.) is the series in which the common difference of two consecutive terms remains the same.