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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] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$

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] ${{a}_{n}}={{S}_{n}}-{{S}_{n-1}}$