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

1.	If the common difference of an ap is 4 than what is a22-a15
Answer» stion :- 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.

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stion :-

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