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Hint: The ${n^{th}}$term of an A.P is ${a_n} = {a_1} + \left( {n - 1} \right)d$. First find the first term and common difference using the two equations, then substitute all values in the sum formula.

Given data

${4^{th}}$Term of an A.P is 22

$ \Rightarrow {a_4} = 22$

And ${15^{th}}$term is 66

$ \Rightarrow {a_{15}} = 66$

Now as we know that the ${n^{th}}$term of an A.P is ${a_n} = {a_1} + \left( {n - 1} \right)d$, where d is the common difference and ${a_1}$ is the first term of an A.P respectively.

$

\Rightarrow {a_4} = {a_1} + \left( {4 - 1} \right)d = {a_1} + 3d = 22 \\

\Rightarrow {a_1} = 22 - 3d.............\left( 1 \right) \\

$

${a_{15}} = {a_1} + \left( {15 - 1} \right)d = {a_1} + 14d = 66$

Now from equation (1) substitute the value of ${a_1}$in above equation we have

$

\Rightarrow 22 - 3d + 14d = 66 \\

\Rightarrow 11d = 66 - 22 = 44 \\

\Rightarrow d = \dfrac{{44}}{{11}} = 4 \\

$

Therefore from equation (1)

$ \Rightarrow {a_1} = 22 - 3\left( 4 \right) = 22 - 12 = 10$

Now as we know that the sum of an A.P is ${S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$, where n is the number of terms.

Now we have to calculate the sum of 8 terms

Therefore $n = 8$

$ \Rightarrow {S_8} = \dfrac{8}{2}\left( {2{a_1} + \left( {8 - 1} \right)d} \right) = 4\left( {2{a_1} + 7d} \right)$

Now substitute the value of ${a_1},{\text{ }}d$in above equation we have

${S_8} = 4\left( {2 \times 10 + 7 \times 4} \right) = 4\left( {20 + 28} \right) = 4\left( {48} \right) = 192$

So, 192 is the required sum of 8 terms of an A.P

Note: In such types of questions the key concept we have to remember is that always recall the basic formulas of A.P which is stated above then according to given conditions and the formulas first calculate the value of first term and common difference of an A.P respectively, then from the formula of sum of an A.P calculate the value of sum of 8 terms and after simplification we will get the required answer.

Given data

${4^{th}}$Term of an A.P is 22

$ \Rightarrow {a_4} = 22$

And ${15^{th}}$term is 66

$ \Rightarrow {a_{15}} = 66$

Now as we know that the ${n^{th}}$term of an A.P is ${a_n} = {a_1} + \left( {n - 1} \right)d$, where d is the common difference and ${a_1}$ is the first term of an A.P respectively.

$

\Rightarrow {a_4} = {a_1} + \left( {4 - 1} \right)d = {a_1} + 3d = 22 \\

\Rightarrow {a_1} = 22 - 3d.............\left( 1 \right) \\

$

${a_{15}} = {a_1} + \left( {15 - 1} \right)d = {a_1} + 14d = 66$

Now from equation (1) substitute the value of ${a_1}$in above equation we have

$

\Rightarrow 22 - 3d + 14d = 66 \\

\Rightarrow 11d = 66 - 22 = 44 \\

\Rightarrow d = \dfrac{{44}}{{11}} = 4 \\

$

Therefore from equation (1)

$ \Rightarrow {a_1} = 22 - 3\left( 4 \right) = 22 - 12 = 10$

Now as we know that the sum of an A.P is ${S_n} = \dfrac{n}{2}\left( {2{a_1} + \left( {n - 1} \right)d} \right)$, where n is the number of terms.

Now we have to calculate the sum of 8 terms

Therefore $n = 8$

$ \Rightarrow {S_8} = \dfrac{8}{2}\left( {2{a_1} + \left( {8 - 1} \right)d} \right) = 4\left( {2{a_1} + 7d} \right)$

Now substitute the value of ${a_1},{\text{ }}d$in above equation we have

${S_8} = 4\left( {2 \times 10 + 7 \times 4} \right) = 4\left( {20 + 28} \right) = 4\left( {48} \right) = 192$

So, 192 is the required sum of 8 terms of an A.P

Note: In such types of questions the key concept we have to remember is that always recall the basic formulas of A.P which is stated above then according to given conditions and the formulas first calculate the value of first term and common difference of an A.P respectively, then from the formula of sum of an A.P calculate the value of sum of 8 terms and after simplification we will get the required answer.