Question

The 10th term of an arithmetic series, S is 66
The sum of the first 20 terms of is S 1290

Answer 1

The question given here is incomplete. The correct question is given below-

"The 10th term of an arithmetic series, S is 66. The sum of the first 20 terms of S is 1290. Find the 5th term of the series."

Given:-

$a_{10}$ = 66

$S_{20}$ = 1290

To find:-

The value of $a_{5}$

Solution:-

We know that,

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

$a_{10}$ = a +(10 - 1)d

    = a + (9)d

66 = a + 9d                                                  (using the value  $a_{10}$ as given)

a = 66 - 9d                                                     (1)

Now,

$S_{20}$ = $\frac{n}{2}$ ×(2a + (n - 1)d)

1290 = $\frac{20}{2}$ ×(2a + ( 20 - 1)d)

1290 = 10 × (2a + 19d)

129 = 2a + 19d

129 = 2 (66 - 9d) + 19d                               (substituting the value of 1 here)

129 = 132 - 18d + 19d

129 - 132 = d

d = -3                                                  (2)

Using the value of d in equation 1,

a = 66 - 9d

a = 66 - 9 (-3)

a = 66 + 27

a = 93

Now,

a₅ = a + (n - 1)d

a₅ = 93 + (5 - 1) × -3

a₅ = 93 -12

a₅ = 81

Hence, $a_{5}$ = 81

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