Question

The sum of the 5th and 9th term or an ap is 40 and the sum of the 8th and 14th term us 64 find the sum of first 20 terms

Answer 1

Let a be the first term of the Arithmetic progression, d be the common difference of the Arithmetic progression.

5th term = a + 4d

9th term = a + 8d.

According to the question,

The sum of 5th and 9th terms of AP is 40.

So, ( a + 4d)+( a + 8d) = 40

2a + 12 d = 40

a + 6d = 20

8th term = a + 7d

14th term = a + 13d

According to the question,

The sum of 8th and 14th terms of AP is 64

So, a + 7d + a + 13 d = 64

2a + 20 d = 64

a + 10d = 32

We have two equations in two variables, We shall now solve them.

a + 10d = 32

a + 6d = 20

We get, 4d = 12, d = 3 on subtracting both the equations.

Now, a = 20 - 6d = 20 - 6(3) = 2

Sum of n terms of an A.P is defined,

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

We need sum of first 20 terms, so n = 20

$S_{20} = \frac{20}{2} (2 \times 2 + (20 - 1) \times 3) \\ \\ \: \: \: \: \: \: \: \: = 10(4 + 19 \times 3) \\ \\ \: \: \: \: \: \: \: \: = 10(4 + 57) = 10 \times 61 = 610$

Therefore, The sum of first 20 terms is 610

Answer 2

Answer:

610

Step-by-step explanation:

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

We have :

Term formula applying it with given sum of term

t_5 + t_9 = 40

$40 = a + 4d + a + 8d \\ \\ 2a + 12d = 40 \\ \\ a = 20 - 6d....(i)$

Also t_8 + t_14 = 64

$64 = a + 7d + a + 13d \\ \\ 2a + 20d = 64 \\ \\ a = 32 - 10d...(ii)$

From ( i ) and ( ii )

20 - 6 d = 32 - 10 d

4 d = 12

d = 3

a = 20 - 6 d

a = 20 - 18

a = 2

Now sum of first 20 term :

Now sum of first 20 term : S_20 = 20 / 2 ( 4 + 19 × 3 )

S_20 = 10 × 61

S_20 = 610

Hence we get answer.