Question

if the 8th term of an ap is 31 and the 15th term is 16 more than the 11th term.
1. find the AP
2. find the sum of first 15 sum​

Answer 1

EXPLANATION.

8th term of an Ap = 31

15th term of am Ap = 16 + 11th term.

Nth term of an Ap.

An = a + ( n - 1 )d.

→ a + 7d = 31 ...........(1)

→ a + 14d = 16 + a + 10d.

→ 4d = 16.

→ d = 4.

Put the value of D = 4 in equation (1) we get,

→ a + 7(4) = 31.

→ a + 28 = 31.

→ a = 3.

First term = a = 3.

Common difference = d = 4.

(1) = To find the Ap.

An = 3 + ( n - 1 ) 4.

An = 3 + 4n - 4.

An = 4n - 1.

Put n = 1 → 4(1) - 1 = 3.

put n = 2 → 4(2) - 1 = 7.

put n = 3 → 4(3) - 1 = 11.

put n = 4 → 4(4) - 1 = 15.

Sequence = 3,7,11,15,.........

(2) = Sum of first 15th term.

Sum of nth term of an Ap.

$\sf \: \implies \: s_{n} \: = \dfrac{n}{2}(2a \: + (n - 1)d) \\ \\ \sf \: \implies \: s_{15} \: = \frac{15}{2} (2 \times 3 + 14 \times 4) \\ \\ \sf \: \implies \: s_{15} = \frac{15}{2} (6 + 56) \\ \\ \sf \: \implies \: s_{15} = \frac{15}{2} \times 62 \\ \\ \sf \: \implies \: s_{15} = 15 \times 31 = 465$

Answer 2

Answer:

Step-by-step explanation:

let a be the 1st term and d be the common difference

a8 = a + 7d = 31  

a15 = a + 14d = a11 + 16

(a + 10d) + 16

a + 10d +16 = a + 14d  

= a + 10d +16 = a + 10d + 4d

= 16 = 4d  

d = 16/4 = 4

a + 7d = 31

a + 7(4) = 31

a + 28 = 31  

a = 31 -28 = 3

a = 3 , d = 4 ...

therefore Ap = a , a+d , a+2d ,a+3d ......

= 3 , 3+4 ,3+2(4) , 3 + 3(4)

= 3 , 7 , 11 , 15 .......

2. S15=  15/2 [ 2[3] + [14] [4] ]

        =15/2 [ 6 + 56]

        =15/2 [ 62

        =15 [ 31]

        =465

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