Question

If the common difference of an ap is 4 than what is a22-a15

Answer 1

Question :-

If the common difference of an ap is 4 than what is $\bf{a_{22} - a_{15}}$

To Find :-

The Difference of the 22th term and the 15th term of the AP.

Given :-

Common Difference = 4.

We know :-

$\boxed{\underline{\bf{a_{n} = a_{1} + (n - 1)d}}}$

Where :-

• $\bf{a_{n}}$ = nth term of the AP.

• $\bf{a_{1}}$ = First term of the AP.

• $\bf{d}$ = Common Difference

Solution :-

Equation (i) :

Given :-

• nth term = $\bf{a_{22}}$

• d = 4

Using the formula and substituting the values in it , we get :-

$:\implies \bf{a_{n} = a_{1} + (n - 1)d}$

$:\implies \bf{a_{22} = a_{1} + (22 - 1) \times 4}$

$:\implies \bf{a_{22} = a_{1} + 21 \times 4}$

$:\implies \bf{a_{22} = a_{1} + 84}$

$\therefore \purple{\bf{a_{22} = a_{1} + 84}}\:\:[Equation.(i)]$

Hence, Eq.(i) is $\bf{a_{22} = a_{1} + 84}$

Equation (ii) :

Given :-

• nth term = $\bf{a_{15}}$

• d = 4

Using the formula and substituting the values in it , we get :-

$:\implies \bf{a_{n} = a_{1} + (n - 1)d}$

$:\implies \bf{a_{15} = a_{1} + (15 - 1) \times 4}$

$:\implies \bf{a_{15} = a_{1} + 14 \times 4}$

$:\implies \bf{a_{15} = a_{1} + 56}$

$\therefore \purple{\bf{a_{15} = a_{1} + 56}}\:\:[Equation.(ii)]$

Hence, Eq.(ii) is $\bf{a_{15} = a_{1} + 56}$

On subtracting Eq.(ii) from Eq.(i) , we get :-

$:\implies \bf{a_{22} - a_{15}}$

$:\implies \bf{a_{22} - a_{15} = (a_{1} + 84) - (a_{1} + 56)}$

$:\implies \bf{a_{22} - a_{15} = a_{1} + 84 - a_{1} - 56}$

$:\implies \bf{a_{22} - a_{15} = \not{a_{1}} + 84 - \not{a_{1}} - 56}$

$:\implies \bf{a_{22} - a_{15} = 84 - 56}$

$:\implies \bf{a_{22} - a_{15} = 28}$

$\therefore \purple{\bf{a_{22} - a_{15} = 28}}$

Hence, the difference between the $\bf{a_{22}}$ and $\bf{a_{15}}$ is 28.