Question

an=25,d=2
Sn = 120 so find the n
and a .​

Answer 1

Answer:

2 possiblities,

a= -13, n=20

a= 15 , n=6

Step-by-step explanation:

an=a+(n-1)d=25

=>a+(n-1)2=25

=>a+2n-2=25

=>a+2n=27

=>a+n+n=27

=>a+n=27-n   .......(1)

Sn=(n/2)(2a+(n-1)d)=120

=>(n/2)(2a+(n-1)2)=120

=>(n/2)x2(a+n-1)=120

=>n(a+n-1)=120  . .......(2)

putting value of (a+n) from (1) in (2), we get

=>n(27-n-1)=120

=>n(26-n)=120

=>26n-n²=120

=>n²-26n+120=0

=>n²-20n-6n+120=0

=>n(n-20)-6(n-20)=0

=>(n-6)(n-20)=0

=>n=6 or n=20

for n=6,

a+2n=27

=>a=27-2n

=>a=27-12

=>a=15

for n=20,

a=27-2n

=>a=27-40

=>a= -13

HOPE IT HELPS,

PLEASE THANK, FOLLOW AND MARK AS BRAINLIEST.

Answer 2

GIVEN :–

$\\ \: \: { \huge{.}} \: \: { \bold{ a_{n} = 25}}$

$\\ \: \: { \huge{.}} \: \: { \bold{ d=2}}$

$\\ \: \: { \huge{.}} \: \: { \bold{ S_{n} = 120}}$

TO FIND :–

$\\ \: \: { \huge{.}} \: \: { \bold{a=?}}$

$\\ \: \: { \huge{.}} \: \: { \bold{n=?}}$

SOLUTION :–

• We know that –

$\\ \dashrightarrow { \bold{ a_{n} = a + (n -1)d}}$

• Put the values –

$\\ \implies { \bold{25= a + (n -1)2}}$

$\\ \implies { \bold{25= a +2n -2}}$

$\\ \implies { \bold{a = 27 - 2n \: \: \: \: - - - - eq.(1)}}$

▪︎ And –

$\\ \dashrightarrow { \bold{ S_{n} = \dfrac{n}{2} \{a + a_n \}}}$

• So –

$\\ \implies { \bold{ 120 = \dfrac{n}{2} \{a +25\}}}$

$\\ \implies { \bold{240 =n \{a +25\}}}$

• Using eq.(1) –

$\\ \implies { \bold{240 =n \{27 - 2n +25\}}}$

$\\ \implies { \bold{240 =n (52 - 2n )}}$

$\\ \implies { \bold{240 =52n - 2 {n}^{2}}}$

$\\ \implies { \bold{2 {n}^{2} - 52n + 240 =0}}$

$\\ \implies { \bold{{n}^{2} - 26n + 120 =0}}$

$\\ \implies { \bold{{n}^{2} - 20n - 6n+ 120 =0}}$

$\\ \implies { \bold{n(n -20) - 6(n - 20) =0}}$

$\\ \implies { \bold{(n - 6)(n -20)=0}}$

$\\ \implies \large{ \boxed{ \bold{n = 6 \: ,\: 20}}}$

• Using eq.(1) –

$\\ \implies \large{ \boxed{ \bold{a = 15 \: ,\: - 13}}}$