2a+22d=-26
2a+ 3d=12
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Answer:
d=-2
a=9
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Answered by
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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.
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