Question

what is the sum of the first 14 terms of an arithmetic sequence if the 5th term is 31 and the 9th term is 59?​

Answer 1

Answer:-

Given:

5th term of an AP = 31

9th term = 59

We know that,

nth term of an AP – a(n) = a + (n - 1)d

Hence,

⟹ a + (5 - 1)d = 31

⟹ a + 4d = 31 -- equation (1)

Similarly,

⟹ a + (9 - 1)d = 59

⟹ a + 8d = 59 -- equation (2)

Subtract equation (1) from (2).

⟹ a + 8d - ( a + 4d ) = 59 - 31

⟹ a + 8d - a - 4d = 28

⟹ 4d = 28

⟹ d = 28/4

⟹ d = 7

Substitute the value of d in equation (1).

⟹ a + 4(7) = 31

⟹ a + 28 = 31

⟹ a = 31 - 28

⟹ a = 3

Now,

Sum of first n terms of an AP – S(n) = n/2 * [ 2a + (n - 1)d ]

⟹ S(14) = 14/2 * [ 2(3) + (14 - 1)(7) ]

⟹ S(14) = 7 [ 6 + 91 ]

⟹ S(14) = 7 * 97

⟹ S(14) = 679

∴ The sum of first 14 terms of the given AP is 679.

Answer 2

GiveN :

• A_5 = 31
• A_9 = 59

To FinD :

• S_14

SolutioN :

⇒A_5 = a + 4d

⇒31 = a + 4d........ (1)

And,

⇒A_9 = 4 + 7d

⇒59 = a + 8d......... (2)

____________________

Solving (1) and (2)

We get,

• a = 3
• d = 7

For Calculating sum of First 14 terms use formula :

⇒Sn = n/2 [2a + (n - 1)d]

⇒S14 = 14/2 [2(3) + (14 - 1)7]

⇒S14 = 7[6 + (13)7]

⇒S14 = 7[6 + 91]

⇒S14 = 7[97]

⇒S14 = 679