Question

If ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}.....$ are in A.P. such that ${{a}_{1}}+{{a}_{5}}+{{a}_{10}}+{{a}_{15}}+{{a}_{20}}+{{a}_{24}}=225$ then ${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+.....+{{a}_{24}}$ is equal to:
(a) 909
(b) 75
(c) 750
(d) 900

Answer 1

Hint: Let us assume the first term of an A.P. is “a” and the common difference of the A.P. is “d”. And we know the general term for an A.P. is equal to: ${{T}_{n}}=a+\left( n-1 \right)d$. Using this relation, the second term is calculated by putting the value of n as 2, the third term is calculated by putting the value of n as 3, and so on. Now, using this relation in the given equation, find the relation between $a\And d$ and using this relation find the sum of the first 24 terms in an A.P. We know that the sum of n terms in an A.P. is equal to:
\[{{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)\].

Complete step by step answer:
In the above problem, following terms are the terms of an A.P.
${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}.....$
We have given the equation of some terms of the given A.P. as follows:
${{a}_{1}}+{{a}_{5}}+{{a}_{10}}+{{a}_{15}}+{{a}_{20}}+{{a}_{24}}=225$ …………. (1)
Now, let us take the first term of an A.P. as “a” and the common difference as “d”. And we know the general term for A.P. as follows:
${{T}_{n}}=a+\left( n-1 \right)d$
After that, we are going to put n as 1 for the first term ${{a}_{1}}$, for second term ${{a}_{2}}$ we are going to put n as 2 and in this way, we are going to find the values of ${{a}_{1}},{{a}_{5}},{{a}_{10}},{{a}_{15}},{{a}_{20}},{{a}_{24}}$.
$\begin{align}
  & {{a}_{1}}=a; \\
 & {{a}_{5}}=a+\left( 5-1 \right)d \\
 & \Rightarrow {{a}_{5}}=a+4d; \\
 & {{a}_{10}}=a+\left( 10-1 \right)d \\
 & \Rightarrow {{a}_{10}}=a+9d; \\
 & {{a}_{15}}=a+\left( 15-1 \right)d \\
 & \Rightarrow {{a}_{15}}=a+14d; \\
 & {{a}_{20}}=a+\left( 20-1 \right)d \\
 & \Rightarrow {{a}_{20}}=a+19d; \\
 & {{a}_{24}}=a+\left( 24-1 \right)d \\
 & \Rightarrow {{a}_{24}}=a+23d \\
\end{align}$
Now, using the above relations in eq. (1) we get,
$\begin{align}
  & a+a+4d+a+9d+a+14d+a+19d+a+23d=225 \\
 & \Rightarrow 6a+69d=225 \\
\end{align}$
Taking 3 as common from the L.H.S of the above equation we get,
$3\left( 2a+23d \right)=225$
Dividing 3 on both the sides we get,
$2a+23d=\dfrac{225}{3}$ ……… (2)
Now, we are asked to find the sum of the first 24 terms of an A.P. We know the formula for summation of n terms of an A.P. as:
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
The number of terms (or value of n is 24) so substituting this value of n in the above equation we get,
$\begin{align}
  & {{S}_{24}}=\dfrac{24}{2}\left( 2a+\left( 24-1 \right)d \right) \\
 & \Rightarrow {{S}_{24}}=12\left( 2a+23d \right) \\
\end{align}$
Using eq. (2) in the above equation we get,
$\begin{align}
  & \Rightarrow {{S}_{24}}=12\left( \dfrac{225}{3} \right) \\
 & \Rightarrow {{S}_{24}}=4\left( 225 \right)=900 \\
\end{align}$

So, the correct answer is “Option d”.

Note: To solve the above problem, you have to have a sound knowledge of writing the general term in an A.P. and also how to write the summation of n terms in an A.P. Also, make sure you won’t make any calculation mistakes in the above problem.