Question

If the sum of n terms of an A.P. is 2n^2 + 5n, then find the 4th term

Answer 1

$\large\green{\underline{\boxed{ a4 = 19}}}$

Given:-

Sn = 2n² + 5n

To find:-

4 th term .

Solution:-

Replacing n by (n - 1) ,we get

S(n -1) = $2 {(n - 1)}^{2} + 5(n - 1)$

$\implies$ $2( {n}^{2} + 1 - 2n) + 5(n - 1)$

$\implies$ $s (n - 1) = 2 {n}^{2} + 2 - 4n + 5n - 5$

$\implies$ $s(n - 1) = 2 {n}^{2} + n - 3$

Now,

$\boxed{\sf\red{ an = Sn - S(n-1) }}$

$\implies$ $an = 2 {n}^{2} + 5n - (2 {n}^{2} + n - 3)$

$\implies$ $an = 2 {n}^{2} + 5n - 2{n}^{2} - n + 3$

$\implies$ $an = 4n + 3$

put n = 4 we get,

$\implies$ $a4 = 4 \times 4 + 3$

$\implies$ $a4 = 19$

hence, the 4 th term of A. P is 19.

Answer 2

Answer:

$\huge{\bold{\mathsf{A_4 = 19 }}}$

Step-by-step explanation:

$\underline{\underline{\bold{\mathfrak{Given:}}}}$

$S_n =2{n}^{2} + 5n$

$\underline{\underline{\bold{\mathfrak{To\:find:}}}}$

$\boxed{\bold{\mathsf{4rth\:term\:of\:the\:A.P\:}}}$

That is: $\bold{\mathsf{A_4 = a + (n-1)d }}$

$\rule{200}{1}$

$\huge{\boxed{\bold{\mathrm{Now\:put\:n\:=\:1\:in\:S_n}}}}$

$\bold{\mathsf{S_1 = 2(1)^{2} + 5(1) = 2 + 5 = 7 }}$

Sum of first term will be first term itself

°•° $\implies{\boxed{\bold{\mathsf{ S_1 = A_1= 7 } }}}$

$\huge{\boxed{\bold{\mathrm{Now\:put\:n\:=\:2\:in\:S_n}}}}$

$\bold{\mathsf{ S_2= 2(2)^{2} + 5(2) = 8 + 10 = 18}}$

$\huge{\boxed{\bold{\mathrm{As\:we\:know\:that\:}}}}$

$\bold{\mathsf{S_2 = A_1 + A_2 }}$

$\bold{\mathsf{ 18 = 7 + A_2 }}$

$\bold{\mathsf{ A_2 = 11 }}$

$\underline{\bold{\mathsf{Now\:to\:find\:d\:(common\:difference)}}}$

Apply

$\bold{\mathsf{A_2 - A_1 = d }}$

$\underline{\bold{\mathsf{Put\:in\:the\:values\:}}}$

$\bold{\mathsf{d= 11 - 7 = 4 }}$

Now put in to get $\bold{\mathsf{ A_4}}$th term of the A.P

$\bold{\mathsf{A_n = a + (n-1)d }}$

$\bold{\mathsf{ A_4 = 7 + (4-1)4}}$

$\bold{\mathsf{A_4 = 7 + 12 = 19 }}$

Hence we get $\bold{\mathsf{A_4 = 19 }}$

$\rule{200}{1}$

$\boxed{\underline{\underline{\bold{\mathfrak{Formula\:Used}}}}}$

★$\bold{\mathsf{A_n = a + (n-1)d }}$

★$\bold{\mathsf{S_1 = A_1 }}$ in any A.P