Find the common difference of an AP whose first term is 5 and the sum of its first
Question:

Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms

Solution:

Let the common difference of the AP be d.

First term, a = 5

Now,

$a_{1}+a_{2}+a_{3}+a_{4}=\frac{1}{2}\left(a_{5}+a_{6}+a_{7}+a_{8}\right)$         (Given)

$\Rightarrow a+(a+d)+(a+2 d)+(a+3 d)=\frac{1}{2}[(a+4 d)+(a+5 d)+(a+6 d)+(a+7 d)] \quad\left[a_{n}=a+(n-1) d\right]$

$\Rightarrow 4 a+6 d=\frac{1}{2}(4 a+22 d)$

$\Rightarrow 8 a+12 d=4 a+22 d$

$\Rightarrow 22 d-12 d=8 a-4 a$

$\Rightarrow 10 d=4 a$

$\Rightarrow d=\frac{2}{5} a$

$\Rightarrow d=\frac{2}{5} \times 5=2 \quad(a=5)$

Hence, the common difference of the AP is 2.

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