Question

The sum of first 25 terms of an Arithmetic sequence is 650 . a) Find its 13th term. b) What is the sum of its 11th and 15th terms? c) Find the sum of its 2 nd and 24th terms?​

Answer 1

Answer:

a) 13th term = 26

b) Sum of 11th and 15th term = 52

c) Sum of 2nd and 24th term = 52

Step-by-step explanation:

We are given,

S(25) = 650

Now we know that,

Sn = (n/2)[2a + (n - 1)d]

Then, we have,

n = 25

Let

a = a

d = d

S(25) = (25/2)[2(a) + (25 - 1)d]

650 = (25/2)[2a + 24d]

650 × 2 = 25[2a + 24d]

1300 = 25[2a + 24d]

1300/25 = 2a + 24d

52 = 2a + 24d ------- 1

52 = 2(a + 12d)

a + 12d = (52/2)

a + 12d = 26 ------ 2

a)

Now,

We know that,

a(nth) = a + (n - 1)d

a(13th) = a + (13 - 1)d

a(13th) = a + 12d

But, from eq.2 we know that,

a + 12d = 26

Thus,

13th term = 26

b)

Similarly,

a(11th) = a + (11 - 1)d

a(11th) = a + 10d ------- 3

and

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

a(15th) = a + 14d ------- 4

Now we need to add the 11th and 15th terms,

So, from eq.3 and eq.4

We have,

Sum = a(11th) + a(15th)

S = (a + 10d) + (a + 14d)

S = a + 10d + a + 14d

S = 2a + 24d

But from eq.1 we get,

S = 2a + 24d = 52

Hence,

Sum of 11th and 15th term = 52

c)

Again, we know that,

a(2nd) = a + (2 - 1)d

a(2nd) = a + d ------ 5

Similarly,

a(24th) = a + (24 - 1)d

a(24th) = a + 23d ------ 6

Then,

Sum = a(2nd) + a(24th)

From eq.5 and eq.6

S = (a + d) + (a + 23d)

S = a + d + a + 23d

S = 2a + 24d

Again from eq.1 we get,

S = 2a + 24d = 52

Hence,

Sum of 2nd and 24th term = 52

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