Question

If the 12th term of an A.P is $-13$ and the sum of the first four terms is 24, what is the sum of the first 10 terms?

Answer 1

Hint: Use relation of ${{n}^{th}}$ term of AP. You get one relation between fixed terms, common difference. Assume first term as a. Now use the sum of n terms formula to find other relation. Now you have 2 variables, 2 equations of a, d variables. Now we can use the substitution method to solve.
$\begin{align}
  & {{t}_{n}}=a+\left( n-1 \right)d \\
 & {{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \\
\end{align}$
Use these formulas to get the relations.

Complete step-by-step answer:
A sequence of numbers such that the difference of any two consecutive numbers is a constant is called an Arithmetic progression. For example the sequence 1, 2, 3, 4 ……… is an arithmetic progression with common difference $=2-1=1$ .
Now, we need to find ${{n}^{th}}$ term of such sequence. Let us have a sequence with first term a and common difference d.
We know the difference between successive terms is d.
So, $\text{second term = first term }+\text{ }d=a+d$
$\text{third term }=\text{ second term }+d=a+2d$
And so on calculating, we get –
\[{{n}^{th}}={{\left( n-1 \right)}^{th}}term+d=a+\left( n-2 \right)d+d=a+\left( n-1 \right)d\] .
First let us assume the first term to be as ‘a’.
Next let us assume a common difference to be as ‘d’.
By basic knowledge of trigonometry we know that –
${{t}_{n}}=a+\left( n-1 \right)d;{{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
Where, ${{t}_{n}}$ represents ${{n}^{th}}$ term of an A.P
${{S}_{n}}$ represents sum of first n terms of an A.P
By using above, we can find 12th term as –
${{t}_{12}}=a+11d$ …………………………. (1)
By using above formula, we can find sum of 4 terms as –
${{S}_{4}}=2\left[ 2a+3d \right]$
By simplifying it, we get the equation of the form:
${{S}_{4}}=4a+6d$ .
First condition given in the question is 12th term is -13.
Substituting this in equation (1), we get it as:
$a+11d=-13$ ……………… (3)
By substituting sum of 4th term as 24, we get it as:
$4a+6d=24$ ………………… (4)
By substituting 11d on both sides of equation (3), we get:
$a=-13-11d$ ………………….. (5)
By substituting this in equation (4) we get:
$4\left( -13-11d \right)+6d=24$
By adding 52 on both sides of equation, we get it as:
$-38d=76$
By dividing with -38 on both sides we get:
$d=-2$ .
By substituting this in equation (5), we get it as:
$a=-13-11\left( -2 \right)=9$
By substituting $a=9,d=-2$ in equation (5), we get:
$4\left( 4 \right)+6\left( -2 \right)=36-12=24$
RHS=LHS. Hence, verified.
Now substituting $a=9,b=-2,n=10$ in ${{S}_{n}}$ formula:
${{S}_{10}}=\dfrac{10}{2}\left[ 2\left( 9 \right)+\left( 9 \right)\left( -2 \right) \right]$
By cancelling the common terms, we get the value as:
${{S}_{10}}=\dfrac{10}{2}\left( 0 \right)=0$ .
Therefore, the sum of the first 10 terms of this A.P will be 0.

Note: Be careful while substituting a, d into formula if you do reverse you get the wrong answer. Remember that verification of solutions must be done to prove that our result is correct. Similarly you can first find x in terms of y and then substitute and continue. Anyways you will get the same result because the value of x, y won’t change.