Question

please answer it correctly.​

Answer 1

Question :-

The sum of first m terms of an AP is (4m²-m) . If its nth term is 107, find the value of n. Also find the 21st term of this AP.

Solution :-

→ First term of AP will be equal to sum of its first term only .  so,

• a₁ = S₁ = 4(1)²-1 = 3

→ sum of first two terms of AP = a₁ + a₂

• a₁ + a₂ = S₂ = 4(2)²-2 = 14

• a₁ + a₂ = 14

• a₂ = 14 - a₁ = 14 - 3 = 11

→ sum of first three terms of AP = a₁+a₂+a₃

• a₁ + a₂ + a₃ = S₃ = 4(3)²-3 = 33

• a₁ + a₂ + a₃ = 33

• a₃ = 33 - a₁ - a₂

• a₃ = 33 - 3 - 11 = 19

→ so, our AP formed is :

  3 , 11 , 19 ....

Now , as we can see

first term of AP , a = 3

common difference , d = 11 - 3 = 8

given nth term of AP  = 107

aₙ = 107

a + ( n - 1 )d = 107

3 + ( n - 1 ) 8 = 107

8 n - 8 = 107 - 3

8 n = 107 - 3 + 8

8 n = 112

n = 14

Hence, the value of n is 14 .

Now we have to find the 21st term of AP

a₂₁ = a + 20 d

     = 3 + 20 ( 8 )

a₂₁ = 163

Hence, the 21st term of AP will be 163.