Question

find the sum of first 16terms of an ap whose 4th and 9th terms are minus 15 and minus 30 respectively​

Answer 1

Answer:

s16=?

a4 = -15. a+3d = -15 ____eq1

a9 = -30. a + 8d = -30-------eq2

eq1/eq2

a+3d=-15

a + 8d = -30

------------------

5d = -15

-- d= -3---

-----------

a= -6

Answer 2

Answer:

- 456

Explanation:

s16 = ?

a4 = - 15

a4 = a+3d (an = a+ (n-1)d)

a+3d = - 15 .... (i)

a9 = - 30

a9 = a+8d (an = a+ (n-1)d)

a+8d = - 30 .... (ii)

Substituting (i) and (ii), we get

a + 8d = - 30

a + 3d = - 15

(-) (-) (+)

____________

5d = - 15

d = -15/5

d = - 3

Putting value of d in eqn (i) we get,

a + 3 ×(-3) = -15

a = -15+9

a= -6

Now,

S16 = n/2 [2a + (n-1)d]

S16 = 16/2 (- 12 - 45)

S16 = 8 × (- 57)

S16 = -456

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