Question

show that a1, a2........... an from an AP where an is defined an =3+4n, find the sum of the first 15 terms​

Answer 1

Solution:-

Given:-

=> aₙ = 3 + 4n

To find

=> Sum of 15th term

Now take

=> aₙ = 3 + 4n

=> when n = 1

a₁ = 3 + 4 ×1 = 7

=> when n = 2

a₂= 3 + 4 × 2 = 11

=> when n = 3

a₃ = 3 + 4 × 3 = 15

=> when n = 4

a₄ = 3 + 4 × 4 = 19

=> The sequences is

=> 7 , 11 , 15 , 19 ,..........

=> Common difference ( d ) = 11 - 7 = 4

=> First term ( a ) = 7

To find S₁₅ =

Formula

=> Sₙ = n/2{ 2a + ( n - 1 )d}

=> 2S₁₅ = 15{ 2 × 7 + ( 15 - 1 ) × 4 }

=> 2S₁₅ = 15{ 14 + 14 × 4 }

=> 2S₁₅ = 15 { 70 }

=> 2S₁₅ = 1050

=> S₁₅ = 1050/2

=> S₁₅ = 525

Answer

=> Sum of 15 term is 525

Answer 2

Step-by-step explanation:

we want to substitute 15 in terms of n

a15=3+4×15

=3+60

=63