Question

If 12th term of an AP is -13 and the sum of first 4 term is 24, what is the sum of first 10 term

Answer 1

Answer:

Let '(a)' be the first term and 'd' be the common difference of the given AP

Sum of first four terms is given as 24

i.e., (a)+(a+d)+(a+2d)+(a+3d)=24

4a+6d=24

2a+3d=12 ____(1)

12th term is given to be -13

i.e., a+(12-1)d= -13

a+11d= -13_____(2)

Solving (1) and (2)

We get a=9 and d= -2

Hence the sum of first 10 terms

S₁₀ = n/2 × [2a+(n-1)d]

= 10/2 × [2×9+9×(-2)]

= 5×9×0

= 0

Hence sum of first 10 terms of given AP is 0

Hope it helps

Answer 2

Answer:

0

Step-by-step explanation:

T12 = -13 = a +11d

2a +22d = -26 ....(1)

S = 2[2a + 3d] = 24

2a + 3d = 12 ....(2)

19d = -38

d = -2

a = -13-11(-2) = -13 + 22 = 9

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

S10 = 5 [18 + 9(-2)] = 90 - 90 = 0