Question

The sum of the 4th and 8th terms of the ap24 and the sum of the 6th and10th term is 44find the first 5trem of an Ap​

Answer 1

Question :

The sum of the 4th and 8th terms of the AP is 24 and the sum of the 6th and the 10th term is 44 , find the first 5 terms of the AP.

Given :

• Sum of 4th term and 8th term of the AP = 24.

• Sum of 6th and 10th term of the AP = 44.

To find :

First 5 terms of the AP.

Solution :

Let the common difference be d and the first term be a.

Case I :

First let us find the nth term of the 4th term and 8th term.

nth term of the 4th term :

We know the formula for nth term i.e,

$\boxed{\bf{t_{n} = a_{1} + (n - 1)d}}$

Where :

• tn = nth term
• a1 = First term
• d = Common Difference
• n = No. of terms

$:\implies \bf{t_{4} = a_{1} + (4 - 1)d}$

$:\implies \bf{t_{4} = a_{1} + 3d}$

$\boxed{\therefore \bf{t_{4} = a_{1} + 3d}}$

Hence the nth term of 4th term is $\bf{t_{4} = a_{1} + 3d}$

nth term of the 8th term :

We know the formula for nth term i.e,

Where :

• tn = nth term
• a1 = First term
• d = Common Difference
• n = No. of terms

$\boxed{\bf{t_{n} = a_{1} + (n - 1)d}}$

$:\implies \bf{t_{8} = a_{1} + (8 - 1)d}$

$:\implies \bf{t_{8} = a_{1} + 7d}$

$\boxed{\therefore \bf{t_{8} = a_{1} + 7d}}$

Hence the nth term of 8th term is $\bf{t_{8} = a_{1} + 7d}$

Now according to the question , the sum of 4th term and 8th term of the AP is 24.i.e,

$\boxed{\bf{t_{4} + t_{8} = 24}}$

Now by substituting the values 4th term and 8th term in the above equation, we get :

$\underline{:\implies \bf{a_{1} + 3d + a_{1} + 7d = 24}}$

$:\implies \bf{2a_{1} + 10d = 24}$

$\boxed{\therefore \bf{2a_{1} + 10d = 24}}$⠀⠀⠀⠀⠀Eq.(i)

Case II :

First let us find the nth term of the 4th term and 8th term.

nth term of the 6th term :

We know the formula for nth term i.e,

$\boxed{\bf{t_{n} = a_{1} + (n - 1)d}}$

Where :

• tn = nth term
• a1 = First term
• d = Common Difference
• n = No. of terms

$:\implies \bf{t_{6} = a_{1} + (6 - 1)d}$

$:\implies \bf{t_{6} = a_{1} + 5d}$

$\boxed{\therefore \bf{t_{6} = a_{1} + 5d}}$

Hence the nth term of 4th term is $\bf{t_{6} = a_{1} + 5d}$

nth term of the 10th term :

We know the formula for nth term i.e,

Where :

• tn = nth term
• a1 = First term
• d = Common Difference
• n = No. of terms

$\boxed{\bf{t_{10} = a_{1} + (n - 1)d}}$

$:\implies \bf{t_{10} = a_{1} + (10 - 1)d}$

$:\implies \bf{t_{10} = a_{1} + 9d}$

$\boxed{\therefore \bf{t_{10} = a_{1} + od}}$

Hence the nth term of 8th term is $\bf{t_{10} = a_{1} + 9d}$

Now according to the question , the sum of 6th term and 10th term of the AP is 44.i.e,

$\boxed{\bf{t_{6} + t_{10} = 44}}$

Now by substituting the values 6th term and 10th term in the above equation, we get :

$\underline{:\implies \bf{a_{1} + 5d + a_{1} + 9d = 44}}$

$:\implies \bf{2a_{1} + 14d = 44}$

$\boxed{\therefore \bf{2a_{1} + 14d = 44}}$⠀⠀⠀⠀⠀Eq.(ii)

Now by subtracting Eq.(ii) from Eq.(i) , we get :

$:\implies \bf{(2a_{1} + 10d) - (2a_{1} + 14d) = 24 - 44}$

$:\implies \bf{2a_{1} + 10d - 2a_{1} - 14d = 24 - 44}$

$:\implies \bf{- 4d = -20}$

$:\implies \bf{\not{- }4d = \not{-} 20}$

$:\implies \bf{4d = 20}$

$:\implies \bf{d = \dfrac{20}{4}}$

$:\implies \bf{d = 5}$

$\boxed{\therefore \bf{d = 5}}$

Hence the common difference of the AP is 5.

Now putting the value of Common Difference (d) in the Eq.(i) , we get :

$:\implies \bf{2a_{1} + 10d = 24}$

$:\implies \bf{2a_{1} + 10(5) = 24}$

$:\implies \bf{2a_{1} + 50 = 24}$

$:\implies \bf{2a_{1} = - 50 + 24}$

$:\implies \bf{2a_{1} = - 26}$

$:\implies \bf{a_{1} = \dfrac{-26}{2}}$

$:\implies \bf{a_{1} = -13}$

$\boxed{\therefore \bf{a_{1} = -13}}$

Hence the First term of the AP is -13.

To find the First 5 terms of the AP :

First term :

First term of the AP is -13.

Second term :

Second term of the AP is (a + d) i.e,

==> -13 + 5

==> -8

Hence the second term of the AP is -8.

Third term :

Third term of the AP is (a + 2d) i.e,

==> -13 + 2(5)

==> -13 + 10

==> -3

Hence the third term of the AP is -3.

Fourth term :

Third term of the AP is (a + 3d) i.e,

==> -13 + 3(5)

==> -13 + 15

==> 2

Hence the fourth term of the AP is 2.

Fifth term :

Third term of the AP is (a + 4d) i.e,

==> -13 + 4(5)

==> -13 + 20

==> 7

Hence the fifth term of the AP is 7.

Thus , the first 5 terms of the AP are , -13,-8,-3,2,7 .