Question

Find the sum of the first 16 terms of an arithmetic sequence where a4=7 and a7= 16. (I need the procedure y Example) please

Answer 1

Answer:

S₁₆ = 328

Step-by-step explanation:

the nth term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

given a₄ = 7 and a₇ = 16 , then

a₁ + 3d = 7 → (1)

a₁ + 6d = 16 → (2)

subtract (1) from (2) term by term to eliminate a₁

0 + 3d = 9

3d = 9 ( divide both sides by 3 )

d = 3

substitute d = 3 into (1) and solve for a₁

a₁ + 3(3) = 7

a₁ + 9 = 7 ( subtract 9 from both sides )

a₁ = - 2

the sum to n terms of an arithmetic sequence is

$S_{n}$ = $\frac{n}{2}$ [ 2a₁ + (n - 1)d ]

with a₁ = - 2 and d = 3 , then

S₁₆ = $\frac{16}{2}$ [ (2 × - 2) + (15 × 3) ]

     = 8(- 4 + 45)

     = 8 × 41

     = 328