Question

the sum of the 4th and 8th term of an ap is 24 and the sum of 6th and 10th term is 44 find the AP and also the sum of first 16 terms of the AP​

Answer 1

Answer:

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Answer 2

Answer

The AP is -13 ,- 8 , - 3 ,2 , 7, 12,..............

The sum upto 16 terms is 392

Given

• The sum of the 4th and 8th term of an AP is 24
• The sum of 6th and 10th term is 44

To Find

• The AP and the sum of its first 16th terms

Formula to be used

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

Solution

Let us consider the first term and the common difference of the AP be a and d respectively.

Taking the 1st condition we have

4th term + 8th term = 24

$\implies a + 3d + a + 7d = 24 \\ \implies2a + 10d = 24 \\ \implies a + 5d = 12 \\ \implies a = 12 - 5d - - - (1)$

Taking the 2nd condition we have

6th term + 10th term = 44

$a + 5d + a + 9d = 44 \\ \implies2a + 14d = 44 \\ \implies a + 7d = 22 - - - > (2)$

Using the value of equation (1) in (2) we have

$12 - 5d + 7d = 22 \\ \implies2d = 10 \\ \implies d = 5$

Putting the value of d in (1)

$a = 12 - 5 \times 5 \\ \implies a = 12 - 25 \\ \implies a = - 13$

Now the sum upto 16 term of the AP

$S _{16} = \frac{16}{2} [ 2 \times ( - 13) + (16 - 1)5] \\ \implies S _{16} = 8 \times ( - 26 + 75) \\ \implies S _{16} = 392$