Question

If $a_1,a_2,a_3.....$ are in $A.P.$ such that $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$.

Find $a_1+a_2+a_3.........+a_{23}+a_{24}$ = ?

Answer 1

Given: $a_1,a_2,.....$ is in A.P.

$a_1=$ first term

d$=$ common difference

We know that the $n^{th}$ term of an AP is given by $a_n=a_1+(n-1)d$

$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$

$\Rightarrow a_1+(a_2+4d)+(a_2+9d)+(a_1+14d)+(a_1+19d)+(a_1+23d)=225$

$\Rightarrow 6a_1+69d=225$ ……$\left(1\right)$

$\Rightarrow 3(2a_1+23d)=225$

$\Rightarrow 2a_1+23d=\frac{225}{3}=75\dots \dots \left(i\right)$

Now, we have to find

$a_1+a_2+a_3......+a_{23}+a_{24}=S_{24}$

$\therefore S_{24}=\frac{24}{2}[2a_1+(24-1\left)d\right]$

$=12[2a_1+23d]$

$=12\times 75=900$ [Using $eq.\left(i\right)$]