Question

The first and the last terms of an A.P are 7 and 49, respectively. If the sum of all its terms is 420, find its common difference.

Answer 1

Hint: Use the fact that in an A.P ${{S}_{n}}=\dfrac{n}{2}\left( a+{{a}_{n}} \right)$ and ${{a}_{n}}=a+(n-1)d$. Solve the two equations to get the value of n (the number of terms) and d (the common difference).

Complete step-by-step answer:
An arithmetic progression is a sequence of numbers in which adjacent numbers differ by a constant factor.
It is represented in the form a, a+d, a+2d, …
a is called the first term of the A.P and d the common difference.
Here it is given that a = 7 ,${{a}_{n}}=49$and ${{S}_{n}}=420$
We know that in an A.P ${{S}_{n}}=\dfrac{n}{2}\left( a+{{a}_{n}} \right)$
Using we get $420=\dfrac{n}{2}\left( 7+49 \right)$
$\Rightarrow 420=\dfrac{n}{2}\times 56$
$\Rightarrow 420=28n$
Dividing both sides by 28, we get
$\dfrac{420}{28}=\dfrac{28n}{28}$
$\Rightarrow 15=n$
Hence n = 15.
We know that in an A.P ${{a}_{n}}=a+\left( n-1 \right)d$
Using we get
$49=7+\left( 15-1 \right)d$
i.e. $49=7+14d$
Subtracting 7 from both sides we get
$49-7=7+14d-7$
$\Rightarrow 42=14d$
Dividing both sides by 14 we get
$\dfrac{42}{14}=\dfrac{14d}{14}$
i.e. 3 = d
Hence d = 3.
Hence the common difference of the A.P = 3.

Note: [a] This question can also be solved using the formula ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$.
Using ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$ we get
$420=\dfrac{n}{2}\left[ 2\times 7+\left( n-1 \right)d \right]\text{ (i)}$
Also ${{a}_{n}}=a+\left( n-1 \right)d$
Using we get
$49=7+\left( n-1 \right)d$
Subtracting 7 from both sides we get
$49-7=7+\left( n-1 \right)d-7$
$\Rightarrow 42=\left( n-1 \right)d\text{ (ii)}$
Substituting the value of $\left( n-1 \right)d$in equation (i) we get
$420=\dfrac{n}{2}\left( 2\times 7+42 \right)$
i.e. $420=\dfrac{n}{2}\left( 14+42 \right)$
$\Rightarrow 420=\dfrac{n}{2}(56)$
$\Rightarrow 420=28n$
Dividing both sides by 28 we get
\[\dfrac{420}{28}=\dfrac{28n}{28}\]
i.e. n = 15.
Substituting the value of n in equation (ii), we get
$42=\left( 15-1 \right)d$
i.e.
$42=14d$
Dividing both sides by 14 we get
$\dfrac{42}{14}=\dfrac{14d}{14}$
i.e. d = 3.
[b] Some of the most important formulae in A.P are:
[1] ${{a}_{n}}=a+\left( n-1 \right)d$
[2] ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
[3] ${{S}_{n}}=\dfrac{n}{2}\left[ 2l-\left( n-1 \right)d \right]$ where l is the last term.
[4] ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$
[5] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$