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solve the all sum
Answers
Answer:
Q2)
a = 18
d = 16-18 = -2
let number of terms be n
sum to n terms = (n/2)(2a + (n-1)d)
given
sum = 0
(n/2)(2*18 + (n-1)*(-2)( = 0
n(36 - 2n + 2) = 0
n(38 - 2n) = 0
n cannot be zero
so, 38 - 2n = 0
2n = 38
n = 19
Q3.)
given,
t14 = 2*t8
t6 = -8
to find S20
let a be the first term and d the common difference
t6 = -8
a + 5d = -8..............(1)
again
t14 = 2*t8
a + 13d = 2(a + 7d)
a + 13d = 2a + 14d
2a - a = 13d - 14d
a = -d
putting in (1)
-d + 5d = -8
4d = -8
d = -2
a = -d = -(-2) = 2
S20 = (20/2)(2*2+(20-1)*(-2))
= 10(4 - 19*2)
= 10(4 - 38)
= 10*(-34)
= -340
Q4)
given
t16 = 5*t3
t10 = 41
to find S14
let a be the first term and d the common difference
a + 9d = 41.........(1)
a + 15d = 5(a + 2d)
a + 15d = 5a + 10d
5a - a = 15d - 10d
4a = 5d
a = 5d/4
putting in (1)
5d/4 + 9d = 41
5d + 36d = 41*4
41d = 41*4
d = 4
a = 5d/4 = 5*4/4 = 5
S15 = (15/2)(2*5 + (15-1)*4)
= (15/2)*(10 + 14*4)
= (15/2)(10 + 56)
= (15/2)*66
= 15*33
= 495.
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