Question

iii) For an A.P., if s10=150 and s9

=126, find t10.​

Answer 1

Solution:-

Given:-

=> S₁₀ = 150

=> S₉ = 126

To find

=> T₁₀

Formula

=> Sₙ = n/2{ 2a + ( n - 1 )d }

Now

=> S₁₀ = 150

=> 150 = 10/2 { 2a + ( 10 - 1 )d}

=> 150 = 5 { 2a + 9d }

=> 30 = 2a + 9d ....(i)eq

=> S₉ = 126

=> 126 = 9/2{ 2a + (9 - 1 )d}

=> 252 = 9 { 2a + 8d }

=> 28 = 2a + 8d ....(ii)eq

Subtract (ii)eq from (i)eq

=> 2a + 9d - 2a - 8d = 30 - 28

=> d = 2

Put the value of d on (i) eq

=> 30 = 2a + 9d ....(i)eq

=> 30 = 2a + 18

=> 30 - 18 = 2a

=> 12 = 2a

=> a = 6

We get

=> First term (a) = 6

=> Common difference = 2

We.have to find

T₁₀

Formula

=> Tₙ = a + ( n - 1 )d

=> T₁₀ = 6 + ( 10 - 1 ) × 2

=> T₁₀ = 6 + 9 × 2

=> T₁₀ = 6 + 18

=> T₁₀ = 24