Question

for an A.p if s10 =150 and s9=126 ,t10 =?​

Answer 1

Answer - :

t₁₀=24

In an A. P

=>S₁₀=126

Where,

a= 1st term

l= last term

Therfore,

We get,

=>10/2(a+t₁₀)=150

=>a+t₁₀=150/5

=>a+t₁₀=30 .......(1)

=>9/2(a+t₉)=126

=>a+t₉=126/9×2

=>a+t₉=14×2

=>a+t₉=28....... (2)

We subtracted equation (1) and( 2)

=>t₁₀ - t₉=2

common difference =d

=>d=2

=>10/2(2a+2(10−1))=150

=>2a+18=30

=>2a=30−18=12

=>a= 12/2

=>a=6

Answer 2

Answer:

24

Step-by-step explanation:

Let the first be a and common difference is d .

According to the question '

S₁₀ = 150 = (10/2) (2a + 9d)

2a + 9d = 30           ( equation 1)

S₉ = 126 = (9/2) ( 2a + 8d)

2a + 8d = 28        ( eqation 2 )

substract equation 2 from equation 1 , we get

a = 6 & d = 2

t₁₀ = a + 9d  = 6 + 9 *2 = 6 + 18 = 24