The sum of the 3rd and the 7th term of an arithmetic progression is 32 and their product is 220 find the sum of the first 21 terms of the arithmetic progression
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Step-by-step explanation:
Given :
a3 + a7 = 32 (1)
a3 + a7 = 220
Find : s21
Ans : a3 = a + ( 3 - 1 ) d
= a + 2d ( 2)
a7 = a + ( 7 -1 ) d
= a + 6d. (3)
Substitute 2 and 3 in 1
a + 2d + a + 6d = 32
2a + 8d = 32
2( a + 4d ) = 32
a + 4d = 16
a = 16 - 4d (4)
(a + 2d) (a + 6d) = 220
Substitute 4
(16 - 4d + 2d ) ( 16 - 4d + 6d ) = 220
(16 - 2d) (16 + 2d) = 220
(16)^2 + (2d)^2 = 220
256 + 4d^2 = 220
4d^2 = 220 - 256
d^2 = - 36 / 4
d= √-9
d= 3 (5)
Substitute 5 in 4
a = 16- 4(3)
a = 16 - 12
a = 4 (6)
Therefore
S21 = n / 2 [ 2a + ( n - 1 ) d ]
= 21 / 2 [ 2 ( 4 ) + (21 - 1 ) 3 ]
= 21/ 2 [ 8 + (20)3 ]
= 21/2 [ 8 + 60 ]
= 21/2 [ 68 ]
= 21 × 34
= 714
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