an ap consists of 37 terms the sum the of the first 3 terms of it is 12 and the sum of its last 3 term is 318 the find the first and last term of the progression
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Step-by-step explanation:
It's Given that
n = 37
Sum of first three terms = 12
12 = (3/2)(2a + (3 - 1)d
2/3 × 12 = 2a + 2d
2 × 4 = 2(a + d)
8/2 = a + d
4 = a + d. …(1)
Sum of last three terms
318 = (3/2)(2a - (3-1)(4-a)
2/3 × 318 = 2a - 2(4-a)
2 × 106 =2(a + a - 4)
106 = 2a - 4
a = 106+4/2
a = 110/2
a = 55
therefore,
d = 4 - a = 4 - 55 = -51
Last term i.e.
a37 = a + (37-1)d
a37 = 55 + 36(-51)
= 55 - 1836
= -1781
First term is 55 and last term is -1781.
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