Question

The sum of 8 terms of an aritjematic sequence is 360 and its 8 term is 66 find the 4 th term​

Answer 1

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Step-by-step explanation:

Let a & d be the first term of the AP and the common difference respectively

ATQ

$S_{8} = 360$

$t_{8} =66$

As we know that

$t_{n} =a + (n - 1)(d)$

$t_{8} =a + (8 - 1)(d)$

$66 =a + 7(d)$

$a = 66 - 7d$                       .................(equation 1)

Now

$S_{n} =\frac{n}{2}[2a + (n - 1)d]$

$S_{8} =\frac{8}{2}[2a + (8 - 1)d]$

$360 = 4[2a + 7d]$

$360 = 4[2(66 - 7d) + 7d]$                         ( from equation 1 )

$\frac{360}{4} = [2(66 - 7d) + 7d]$

$\frac{360}{4} = [132 - 14d + 7d]$

$90 = 132 - 7d$

$90 - 132 = - 7d$

$- 42 = - 7d$

$42 = 7d$

$d = \frac{42}{7}$

d = 6

Now the common difference = 6

putting the value of d in the equation 1

$a = 66 - 7(6)$

$a = 66 - 42$

$a = 24$

a)

$t_{n} =a + (n - 1)(d)$

$t_{4} =24 + (4 - 1)(6)$

$t_{4} =24 + (3)(6)$

$t_{4} =24 + 18$

$t_{4} = 42$

c)  Every AP is in the for :-

a ; a + d ; a + 2d ; a + 3d.............

d)

$n = 10\\a = 24\\d = 6$

$S_{n} =\frac{n}{2}[2a + (n - 1)d]$

$S_{10} =\frac{10}{2}[2(24) + (10 - 1)(6)]$

$S_{10} =5[ 48 + 54]$

$S_{10} =5[102]$

$S_{10} =510$

Therefore the sum of first 10 terms of the AP is 510