Question

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

Find the 5th term of .

Answer 1

Given :

• 10th term of A.P., a₁₀ = 66
• Sum of first 20 terms, S₂₀ = 1290

To Find :

• 5th term of A.P. = ?

Solution :

We have :

• 10th term of A.P., a₁₀ = 66

Now, we know that :

$\large \underline{\boxed{\bf{a_{n} = a + (n-1)d}}}$

Here,

• $\sf a_{n} = nth\: term\: of \: A.P.$
• a = first term of A.P.
• d = common difference
• n = no. of terms

So, For 10th term of A.P. :

$\sf : \implies a_{10} = a + (10-1)d$

$\sf : \implies a_{10} = a + 9d$

Now, a₁₀ = 66.

$\sf : \implies 66 = a + 9d$

$\sf : \implies a + 9d = 66 \: - (i)$

Now, we are given :

• Sum of first 20 terms, S₂₀ = 1290

We know that :

$\large \underline{\boxed{\bf{S_{n} = \dfrac{n}{2}[2a+(n-1)d]}}}$

Here,

• $\sf S_{n} = sum\: of\: first\: n\: terms\: of\: A.P.$
• a = first term of A.P.
• d = common difference
• n = no. of terms

So, for sum of first 20 terms of A.P. :

$\sf : \implies S_{20} = \dfrac{20}{2}[2a+(20-1)d]$

$\sf : \implies S_{20} = 10 [2a+ 19d]$

Now, S₂₀ = 1290.

$\sf : \implies 1290 = 10 [2a+ 19d]$

$\sf : \implies \dfrac{1290}{10} = 2a+ 19d$

$\sf : \implies 129 = 2a+ 19d$

$\sf : \implies 2a+ 19d = 129 \: - (ii)$

From equation (i), we have :

$\sf : \implies a = 66 - 9d$

By substituting it in equation (ii), we get :

$\sf : \implies 2(66 - 9d) + 19d = 129$

$\sf : \implies 132 - 18d + 19d = 129$

$\sf : \implies 132 + d = 129$

$\sf : \implies d = 129 - 132$

$\sf : \implies d = - 3$

By filling this value in equation (i), we get :

$\sf : \implies a + 9(-3) = 66$

$\sf : \implies a - 27 = 66$

$\sf : \implies a = 66 + 27$

$\sf : \implies a = 93$

Hence, we have :

• First term, a = 93
• Common difference, d = - 3

Now, we have to find 5th term of A.P. :

We know that :

$\large \underline{\boxed{\bf{a_{n} = a + (n-1)d}}}$

$\sf : \implies a_{5} = a + (5-1)d$

$\sf : \implies a_{5} = a + 4d$

By filling values of a and d :

$\sf : \implies a_{5} = 93 + 4(-3)$

$\sf : \implies a_{5} = 81$

$\large \underline{\boxed{\bf{a_{5} = 81}}}$

Hence, 5th term of given A.P. is 81.

Answer 2

Given:-

• a₁₀ = 66
• S₂₀ = 1290

To Find:-

• a₅ = ?

Solution :-

aₙ = a + (n-1)d

a₁₀ = a + (n-1)d

66 = a + (10-1)d

66 = a + 9d

a = 66-9d ... [eq.-1]

Sₙ = n/2 × [2a+(n-1)d]

1290 = 20/2 [2a + (20-1)d

1290 = 20/2 [2a + 19d]

129 = 2a + 19d ...eq(2)

Now, substitute value of a in equation 2,

129 = 2(66-9d) + 19d

129 = 132 - 18d + 19d

129 = 132 + d

d = 129-132

d = -3

By substituting value of d in equation 1,

a = 66 -9d

a = 66-9(-3)

a = 93

Now,

a₅ = a + (n-1)d

a₅ = 93 + (5-1)-3

a₅ = 93 -12

a₅ = 81