Question

Find n for the arithmetic sequence for which sn=345, u1=12 and d = 5 .

Answer 1

Answer:

n = 10

Step-by-step explanation:

The sum to n terms of an arithmetic sequence is

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

where a₁ is the first term and d the common difference

Here a₁ = 12 and d = 5 and $S_{n}$ = 345, thus

$\frac{n}{2}$ [ (2 × 12) + 5(n - 1) ] = 345 ( multiply both sides by 2 )

n( 24 + 5n - 5) = 690 ← distribute and simplify left side

n(19 + 5n) = 690

19n + 5n² = 690 ( subtract 690 from both sides )

5n² + 19n - 690 = 0 ← in standard form

(5n + 69)(n - 10) = 0 ← in factored form

Equate each factor to zero and solve for n

5n + 69 = 0 ⇒ 5n = - 69 ⇒ n = - $\frac{69}{5}$

n - 10 = 0 ⇒ n = 10

However, n > 0 , thus n = 10