Question

Find the sum of first 17 term of ap whose 4th and 9th term are - 15 and minus 13 respectively

Answer 1

Hey friend, your answer is -
Here, t4 = -15 and t9 = -13
tn = a + (n -1 ) d.........formula
t4 = a + 3d
a + 3d = -15......... (1)

Similarly, a + 8d = -13.......... (2)
Subtract equation 1 from 2 ,
a + 8d = -13
-
a + 3d = -15
------------------
5d = 2
d = 2 /5

Substitute d= 0.4 in equation 1 ,
a + 3×2/5 = -15
5a + 6 = -75
5a = -81
a = -81 / 5

Sn = n/2 (2a + (n-1)d)
S17 = 17 / 2 (2× -81 /5 +(17-1)×2/5)
= 17/2 (-162 /5 + 32 / 5
= 17 / 2 × -130 / 5
= - 442 / 2
= - 221

S17 = - 221

Hope it will help you.

Answer 2

a4 = -15 (given)

a9 = -13 (given)

since, a4 = a + (n-1)d_(given that a4 = -15)

-15= a + (4-1)d

a + 3d = -15 ______ (i)

and, a9 = a + (n-1)d__(given that a9 = -13)

-13= a + (9-1)d

a + 8d = -13 _______(ii)

by eliminating eq (i) and (ii)

a + 3d = -15

a + 8d = -13

- - +

_____________

-5d = -2

d = 2/5

put the value of d in eq (i)

a + 3(2/5) = -15

a + 6/5 = -15

a = -15 - 6/5

a = (-75-6)/5

a = -81/5

Now, to find the sum of 17 term, we have

a = -81/5

d = 2/5

and, n = 17

So, s17 = n/2 (2a + (n-1)d)
= 17/2 (2(-81/5)+(16)2/5)
= 17/2 (-162/5)+32/5)
= 17/2 (-130/5)
= -221
Hence, sum if the 17 terms of the AP is -221