How many terms must be taken of the series 42,39,36...... to make the sum 315?
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Answered by
16
here ,
a = t1 = 42
t2 = 39
so , d = t2 - t1 = 39 - 42
= - 3
Sn = 315
find - n = ?
Sn = n/2 [ 2a + ( n-1) d]
315 = n/2 [ 2 × 42 + ( n-1) -3]
630 = n [ 84 - 3n + 3]
630 = n [ 87 - 3n]
630 = 87n - 3n^2
3n^2 - 87n + 630 = 0
n^2 - 29n + 210 = 0
n^2 - 15n - 14n + 210 = 0
n(n-15) - 14( n-15) = 0
(n-15) ( n-14) = 0
(n-15) =0 or ( n-14) = 0
n = 15 or n = 14
42 , 39 , 36 , .......3 , 0
if taking 0 in account , n = 15
if not n = 14
a = t1 = 42
t2 = 39
so , d = t2 - t1 = 39 - 42
= - 3
Sn = 315
find - n = ?
Sn = n/2 [ 2a + ( n-1) d]
315 = n/2 [ 2 × 42 + ( n-1) -3]
630 = n [ 84 - 3n + 3]
630 = n [ 87 - 3n]
630 = 87n - 3n^2
3n^2 - 87n + 630 = 0
n^2 - 29n + 210 = 0
n^2 - 15n - 14n + 210 = 0
n(n-15) - 14( n-15) = 0
(n-15) ( n-14) = 0
(n-15) =0 or ( n-14) = 0
n = 15 or n = 14
42 , 39 , 36 , .......3 , 0
if taking 0 in account , n = 15
if not n = 14
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