Question

in an A.P. 19th term is 52 and 38th term 128,find sum of first 56 term.

Answer 1

a = -20
d = 4

&
Sum of 56 terms will be 5040

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Answer 2

:

$\huge\bold\orange{➢ANSWER :-}$

➢For an A.P., let a be the first term and d be the common difference.

$==>> t19 = 52, t38 = 128 …[Given]$

$==>>Since, tn = a + (n – 1)d$

∴

$t19 = a + (19 – 1)d$

$∴ 52 = a + 18d i.e. a + 18d = 52 …(i)$

Also, t38 = a + (38 – 1)d

∴

$128 = a + 37d \\ . …(ii)$

➢ Adding equations (i) and (ii), we get

$==>> a + 18d = 52$

$(a + 37d = 128)/(2a + 55d = 180) .....(iii)$

➢ Now,

$Sn = n/2 [ 2a + (n - 1)d] \\ </p><p>∴ S56 = 56/2 [ 2a + (56 - 1)d] \\ </p><p>= 28(2a + 55d) = 28 x 180 ....[From (iii) ] \\ </p><p>∴ S56 = 5040$

∴ The sum of the first 56 terms is 5040