Question

An AP consists of three terms whose sum is 15 and sum of their squares of extremes is 58. Find the first three terms of AP and also find the sum of first 50 terms of an Ap

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

The sum of first 50 terms of an A.P is 2700 .

Step-by-step explanation:

Given as :

For an A.P , the sum of three terms = 15

Let The three A.P terms are,  a - d  ,  a  , a + d

So, The sum of three terms = (a -d) + a + (a + d)

Or,  (a -d) + a + (a + d) = 15

Or, (a + a + a) + ( - d + d) = 15

Or, 3 a + 0 = 15

∴  a = $\dfrac{15}{3}$

i.e  a = 5

So, The first term = a = 5

Again

The sum of square of extremes = 58

So, (a - d)² + (a + d)² = 58

Or, a² - 2 a d + d² + a² + 2 a d + d² = 58

Or, 2 a² + 2 d² + 0 = 58

Or, a² + d² = $\dfrac{58}{2}$

Or, a² + d²  = 29

Put the value of a

So, 5² + d²  = 29

Or, 25 + d²  = 29

Or,  d²  = 29 - 25

Or, d²  = 4

∴  d = $\sqrt{4}$

i.e d = 2

So, The common difference in A.P = d = 2

∴ The first term of A.P = a - d = 5 - 2

I.e The first term of A.P = 3

The second term of A.P = a

i.e The second term of A.P = 5

The third term of A.P = a + d = 5 + 2

i.e The third term of A.P = 7

Again

Let The sum of first 50 terms = s

∵ The sum of n terms of A.P = $\dfrac{n}{2}$ [2 a + (n - 1) d]

So, The sum of first 50 terms = $\dfrac{50}{2}$ [2 × 5 + (50 - 1) 2]

Or, s = 25 × [10 + 49 × 2]

Or, s = 25 × [108]

Or, s = 2700

So, The sum of first 50 terms of an A.P = s = 2700

Hence, The sum of first 50 terms of an A.P is 2700 . Answer