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according to question
T3 = 20 & T8 = 10
let first term of A.p. is a and difference d
now
T3 = a + 2d = 20 ....(1)
T8 = a + 7d = 10 .....(2)
—
———————————————
-5d = 10
d = -2
———————————————
now put value of d in equation (1)
a - 4 = 20
a = 24 ...(3)
sum of terms = 124
Sn = n/2[2a + (n-1)d]
124 = n/2 [2×24 + (n-1)(-2) ]
248 = n [ 48 -2n +2]
248 = 48n - 2n² + 2
2n² - 48n + 246 = 0
n² - 24n + 123 = 0
solve it & get value of n ..
hope it will help you....✌✌
T3 = 20 & T8 = 10
let first term of A.p. is a and difference d
now
T3 = a + 2d = 20 ....(1)
T8 = a + 7d = 10 .....(2)
—
———————————————
-5d = 10
d = -2
———————————————
now put value of d in equation (1)
a - 4 = 20
a = 24 ...(3)
sum of terms = 124
Sn = n/2[2a + (n-1)d]
124 = n/2 [2×24 + (n-1)(-2) ]
248 = n [ 48 -2n +2]
248 = 48n - 2n² + 2
2n² - 48n + 246 = 0
n² - 24n + 123 = 0
solve it & get value of n ..
hope it will help you....✌✌
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