Question

In an AP of 50 terms, the sum of the first 10 terms is 210 and the sum of its last 15 terms is 2565. Find the AP.

Answer 1

Hint: Assume that the first term of the AP is ‘a’ and the common difference is ‘d’. Write the last term of the AP using the formula ${{a}_{n}}=a+\left( n-1 \right)d$ for writing the ${{n}^{th}}$ term of the AP. To write the sum of first ‘n’ terms of the AP, use the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$. Write equations based on the data given in the question. Simplify the equations to calculate the values of variables ‘a’ and ‘d’ and thus, find the AP.

Complete step-by-step answer:
We have to find the AP of 50 terms such that the sum of the first 10 terms is 210 and the sum of its last 15 terms is 2565.
Let’s assume that the first term of the AP is ‘a’ and the common difference is ‘d’.
We know that we can write the ${{n}^{th}}$ term of the AP using the formula ${{a}_{n}}=a+\left( n-1 \right)d$, where ‘a’ is the first term of the AP and ‘d’ is the common difference.
Substituting $n=50$ in the above formula, we have ${{a}_{50}}=a+\left( 50-1 \right)d=a+49d$.
We know that the sum of first ‘n’ terms of the AP is ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, where ‘a’ is the first term of the AP and ‘d’ is the common difference.
Substituting $n=10$ in the above formula, we have ${{S}_{10}}=\dfrac{10}{2}\left[ 2a+\left( 10-1 \right)d \right]=5\left[ 2a+9d \right]$.
We know that the sum of the first ‘n’ terms is 210. Thus, we have $5\left( 2a+9d \right)=210$.
Simplifying the above equation, we have $2a+9d=\dfrac{210}{5}=42.....\left( 1 \right)$.
We will now consider the reverse of the given AP, i.e., it’s first term is the ${{50}^{th}}$ term of the given AP, the second term is the ${{49}^{th}}$ term of the given AP, and so on. We observe that the common difference of this AP is –d and its first term is ${{a}_{50}}=a+49d$.
We know that the sum of the first 15 terms of this AP is 2565. Substituting $n=50$ in the formula ${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$, we have ${{S}_{15}}=\dfrac{15}{2}\left[ 2\left( a+49d \right)+\left( 15-1 \right)\left( -d \right) \right]=\dfrac{15}{2}\left[ 2a+98d-14d \right]=\dfrac{15}{2}\left( 2a+84d \right)=15\left( a+42d \right)$.
We know that this sum is equal to 2565. Thus, we have $15\left( a+42d \right)=2565$.
Simplifying the above equation, we have $a+42d=\dfrac{2565}{15}=171.....\left( 2 \right)$.
We will now solve equation (1) and (2) by the elimination method. Multiplying equation (2) by 2 and subtracting equation (1) from it, we have $2\left( a+42d \right)-2a+9d=2\times 171-42$.
Simplifying the above equation, we have $84d-9d=342-42\Rightarrow 75d=700$.
Thus, we have $d=\dfrac{300}{75}=4$.
Substituting $d=4$ in equation (2), we have $a+42\left( 4 \right)=171$.
Simplifying the above equation, we have $a+168=171$.
Thus, we have $a=171-168=3$.
Hence, the AP is $3,3+4,3+2\left( 4 \right),...=3,7,11,...$.

Note: Arithmetic Progression (AP) is a sequence of numbers such that the difference between any two terms is a constant. It’s necessary to use the fact that when the AP is reversed, the first term of the new AP is equal to the last term of the given AP and the common difference of the new AP is negative of the common difference of the given AP.