Question

2a+22d=-26
2a+ 3d=12​

Answer 1

Answer:

d=-2

a=9

good night dear

Answer 2

Answer:

Let a be the first term and d be the common difference.

nth tem of an AP is given by,

a(n) = a + (n - 1)d

and

Sum of n terms is given by,

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

Therefore, we have,

a + 11d = -13 ... (1)

and,

sum of 4 terms = 24

4/2 [2a + (n - 1)d] = 24

2 [2a + (n - 1)d] = 24

2[2a + (4 - 1)d = 24

2[2a + 3d] = 24

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

Multiplying eq. (1) by 2 we get,

2a + 22d = -26 ... (3)

Now,

Subtratcing eq. (2) by eq. (3), w get,

19d = -38

d = -2

Substituting the value of d in eq. (1), we get,

a + 11(-2) = - 13

a - 22 = -13

a = -13 + 22

a = 9

Now, sum of first 10 terms is,

s(10) = 10/2 [2(9) + (10 - 1)(-2)]

= 10/2 [18 - 18]

= 5 (0)

= 0

Therefore, sum of first 10 terms of the given AP is 0.