Question

In an A.P. the sum of the first ten terms is -150 and the sum of its next ten terms is -550. Find first term of A.P.

Answer 1

Hint: In the above equation we will use the formula of the sum of the first n terms of an A.P. which is as follows:
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Where n is the number of terms, a is the first term of the A.P. and ‘d’ is the common difference.

Complete step-by-step answer:

We have been given the sum of the first ten terms to be -150 and the sum of its next ten terms to be -550.
Let us suppose the first term to be ‘a’ and the common difference to be ‘d’ of the A.P.
Since we know that the sum of the first n terms of an A.P. is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$ where ‘a’ is the first term and ‘d’ is common difference.
We have n=10 and ${{S}_{n}}=-150$.

& \Rightarrow -150=\dfrac{10}{2}\left[ 2a+\left( 10-1 \right)d \right] \\\ & \Rightarrow -150=5\left[ 2a+9d \right] \\\ & \Rightarrow \dfrac{-150}{5}=2a+9d \\\ & \Rightarrow -30=2a+9d.....(1) \\\ \end{aligned}$$ Now the sum of the next ten terms is -550. So we have n=10 and $${{S}_{n}}=-550$$ Here, the first term is equal to the 11th term of the A.P. $$\Rightarrow a'=a+\left( 11d-1 \right)d=a+10d$$ Since we know that the $${{n}^{th}}$$ term is given by $${{T}_{n}}=a+\left( n-1 \right)d$$ for an A.P. having first term ‘a’ and common difference ‘d’. $$\begin{aligned} & -550=\dfrac{10}{2}\left[ 2\left( a+10d \right)+\left( 10-1 \right)d \right] \\\ & -550=5\left[ 2a+20d+9d \right] \\\ & \dfrac{-550}{5}=\left( 2a+29d \right) \\\ & -110=2a+29d......(2) \\\ \end{aligned}$$ Subtracting equation (1) from (2), we get as follows: $$\begin{aligned} & -110-(-30)=2a+29d-2a-9d \\\ & -80=20d \\\ & \Rightarrow \dfrac{-80}{20}=d \\\ & \Rightarrow -4=d \\\ \end{aligned}$$ On substituting the value of ‘d’ in equation (1) we get as follows: $$\begin{aligned} & -30=2a+9(-4) \\\ & -30=2a-36 \\\ & -30+36=2a \\\ & 6=2a \\\ & \Rightarrow a=\dfrac{6}{2} \\\ & \Rightarrow a=3 \\\ \end{aligned}$$ Therefore, the first term of the A.P. is equal to 3. Note: Be careful while doing the calculation and also take care of the sign during calculation. Remember that arithmetic progression is a sequence of numbers in order that the difference of any two successive numbers is a constant value, it is denoted as A.P.