Question

If the sum of first 4 terms of an AP is 40 and that of first 14 terms is 280, find the sum of first $n$ terms.

Answer 1

Put the values given in the question in the sum of first $n$ terms formula of any AP with common difference $d$ and the first term $a$ that is $S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ to obtain two linear equations. Solve them to get $d$, $a$ and then $S$.

Complete step-by-step answer:

Arithmetic sequence otherwise known as arithmetic progression, abbreviate d as AP is a type numerical sequence where the difference between any two consecutive numbers is constant . If $\left( {{x}_{n}} \right)={{x}_{1}},{{x}_{2}},{{x}_{3}},...$ is an AP with infinite terms, then ${{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$ . The difference between two terms is called common difference and denoted $d$ where $d={{x}_{2}}-{{x}_{1}}={{x}_{3}}-{{x}_{2}}...$. The term ${{x}_{1}}$ is called appearing first in the sequence is called the first term.

We know that the sum of first $n$ of terms of any AP with common difference $d$ and the first term $a$ is given by

$S=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$

It is given that in AP which is given the question that the sum of first 4 terms is 40. So putting $n=4$ and $S=40$ in above we get,

$40=\dfrac{4}{2}\left( 2a+\left( 4-1 \right)d \right)$

$\Rightarrow 2a+3d=20...\left( 1 \right)$

It is given that in AP which is given the question that the sum of first 14 terms is 280. So putting $n=4$ and $S=40$ in above we get,

$280=\dfrac{14}{2}\left( 2a+\left( 14-1 \right)d \right)$

$\Rightarrow 2a+13d=40...\left( 2 \right)$

We solve the linear equations (1) and (2) by subtracting eqauion (1) from equation (2) to get $d$ and then put in equation to get $a$,

$10d=20\Rightarrow d=2$

$2a=20-3\times 2=14\Rightarrow a=7$

Now we put $a$ and $d$in sum of terms formula

$S=\dfrac{n}{2}\left( 2\times 7+\left( n-1 \right)2 \right)=n\left( 7+n-1 \right)={{n}^{2}}+6n$ which is the required sum.

Note: The difference between sequence and a series is that in series the terms added . We need to care of the sum of terms of AP from GP which has sum of $n$ terms as $S=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$. We cannot find the sum of all the terms of an infinite AP sequence because an AP sequence does not converge.