Question

Find the sum to n terms of arithmetic progressions whose forth and fifth terms are 13 and 15.

Answer 1

Step-by-step explanation:

a +4d=15-------(1)

a+3d=13---------(2)

by subtracting eqn (1) and (2)

d=2

by putting the value of d in eqn 1

a +8 =15

a=7

sum of n terms = n/2 [ 2a + (n-1)d]

=n/2[14+2n-2]

= n/2[12+2n]

= n[6+n]

=n²+ 6n

Answer 2

Given :-

In an arithmetic progression

• fourth term = a₄ = 13
• fifth term = a₅ = 15

To find :-

• sum of n terms of AP

Knowledge required :-

• nth term of AP is given by a + ( n - 1 ) d

( where a is the first term and d is the common difference)

• sum of n term is given by

n( 2a + ( n -1)d ) /2

( where a is the first term and d is common difference)

Solution :-

As we know common difference is given by the difference between any two consecutive terms of an AP

Also,

Since a₄ and a₅ are consecutive terms

therefore,

=> common difference, d = a₅ - a₄

=> d = 15 - 13 = 2

=> d = 2

Now,

=> a₄ = a + (4-1)d = a + 3 d

=> 13 = a + 3 (2)

=> a = 7

Now we have,

first term of AP, a = 7

common difference, d = 2

Now finding the sum of n terms

=> Sum of n terms = n(2a + (n-1)d)/2

=> Sum of n terms = n(2(7) +(n-1)(2))/2

=> Sum of n terms = n(14+2n-2) /2

=> Sum of n terms = n(12+2n) /2

=> Sum of n terms = n (6+n)

=> Sum of n terms = 6 n + n² Ans.

Hence sum of n terms will be given by

6 n + n²