Question

What is the sum of the first 24 terms of the AP $a_1,a_2,a_3,\dots \dots$, if it is known that $a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$? [4 MARKS]

Answer 1

Formula: 1 Mark
Application: 2 Marks
Answer: 1 Mark

We know that in an AP the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of the first and last term.

i.e. $a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\dots$

So, if an A.P consists of 24 terms, then

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

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

$\Rightarrow (a_1+a_{24})+(a_5+a_{20})+(a_{10}+a_{15})=225$

$\Rightarrow (a_1+a_{24})=\frac{225}{3}=75$

$\Rightarrow S_{24}=\frac{24}{2}(a_1+a_{24})$

$[\because S_n=\frac{n}{2}(a_1+a_n\left)\right]$

$S_{24}=12\left(75\right)$

$S_{24}=900$