Question

sum of first n terms is 3n^{2} /2 +5n/2 .find its first term common difference and its 25th term

Answer 1

GIVEN :–

• Sum of n terms $\: { \bold{ (S_{n}) = \dfrac{3 {n}^{2} }{2} + \dfrac{5n}{2} }}$

TO FIND :–

• First term (a) = ?

• Common difference(d) = ?

• 25th term = ?

$\\ \rule{220}{2}$

SOLUTION :–

• We know that nth term of A.P. is –

$\\ \bigstar \: \large \: \: { \green{ \boxed{ \bold{ T_{n} = a + (n - 1)d }}}}$

• So that –

$\\ \longrightarrow \: \: { \bold{ T_{1} = a }}$

$\\ \longrightarrow \: \: { \bold{ T_{2} = a + d }}$

▪︎ Now using –

$\\ \implies { \bold{ S_{n} = \dfrac{3 {n}^{2} }{2} + \dfrac{5n}{2} }}$

• Put n = 1 –

$\\ \implies { \bold{ S_{1} = T_{1} = a = \dfrac{3 {(1)}^{2} }{2} + \dfrac{5(1)}{2} }}$

$\\ \implies { \bold{ S_{1} = T_{1} = a = 4 -----eq.(1)}}$

$\\ \implies \large { \red{ \boxed{ \bold{ a = 4 }}}}$

• Put n = 2 –

$\\ \implies { \bold{ S_{2} = T_{1} + T_{2} = \dfrac{3 {(2)}^{2} }{2} + \dfrac{5(2)}{2} }}$

$\\ \implies { \bold{ S_{2} = T_{1} + T_{2} = 6 \: + 5 = 11}}$

• Put n = 3 –

$\\ \implies { \bold{ S_{3} = T_{1} + T_{2} + T_{3} = \dfrac{3 {(3)}^{2} }{2} + \dfrac{5(3)}{2} }}$

$\\ \implies { \bold{ S_{3} = T_{1} + T_{2} + T_{3} = \dfrac{27}{2} + \dfrac{15}{2} = 21}}$

• Now let's calculate –

$\\ \implies { \bold{ S_{2} - S_{1} = ( T_{2} + T_{1}) - (T_{1})}}$

$\\ \implies { \bold{ 11 - 4 = T_{2} }}$

$\\ \implies { \bold{ T_{2} = 7 }}$

$\\ \implies { \bold{a + d = 7 \: \: \: \: \: \: [ \: \because \: \: T_{2} = a + d]}}$

$\\ \implies { \bold{4 + d = 7 \: \: \: \: \: \: [ \: using \: \: eq.(1)]}}$

$\\ \implies \large { \red{ \boxed{ \bold{ d = 3}}}}$

▪︎ Now 25th term –

$\\ \implies\: \: { \bold{ T_{25} = a + (25 - 1)d }}$

$\\ \implies\: \: { \bold{ T_{25} = 4 + (25 - 1)3 }}$

$\\ \implies\: \: { \bold{ T_{25} = 4 + 24 \times 3 }}$

$\\ \implies\: \: { \bold{ T_{25} = 4 + 72}}$

$\\ \implies\: \: \large { \red{ \boxed{ \bold{ T_{25} = 76}}}}$

$\\ \rule{220}{2}$

Answer 2

plz refer to this attachment