35. An arithmetic progression consists of 37 terms. The sum of the first 3 terms of
it is 12 and the sum of its last 3 terms is 318, then find the first and last terms
of the progression
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Step-by-step explanation:
Let the A.P. be:
a, a+d, a+2d, a+3d ... a+(36-1)d, a + (37-1)d
According to question,
a + a + d + a + 2d = 12
=> 3a + 3d = 12
=> a + d = 4
=> d = 4 - a --- 1
Also,
a + 34d + a + 35d + a + 36d = 318
=> 3a + 105d = 318
=> a + 35d = 106
From equation 1
=> a + 35(4-a) = 106
=> 35×4 - 35a + a = 106
=> 140 - 34a = 106
=> 34a = 34
=> a = 1
d = 4 - a = 3
Therefore,
First term is a = 1
And last term is a + 36d = 1 + 36×3 = 109
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