Question

If square of the second term of an AP is equal to it's 12th term and 10th term is thrice the third term. Find the sum of first 25 terms of that ap

Answer 1

The sum of first 25 terms of that a.p. is 675 .

Step-by-step explanation:

Step 1:  

Let the first term in an A.P. be "a" and the common difference be "d".

We have the formula for the nth term of an A.P. as,

aₙ = a + (n-1)d  

It is given that,  

10th term of an A.P. is thrice the third term of that A.P.  

i.e., a₁₀ = 3 × a₃

⇒ a + (10 - 1)d = 3 × [a + (3-1)d]

⇒a + 9d = 3a + 6d

⇒3d = 2a

⇒ d =  $\frac{2}{3}$ a ......... (i)

Step 2:

Also, square of the 2nd term of an A.P. is equal to its 12th term

i.e., a₂² = a₁₂

⇒ [a + d]² = a + 11d

substituting from (i), we get

⇒ [a + $\frac{2}{3}$ a]² = a + 11× $\frac{2}{3}$ a

⇒  [$\frac{5a}{3}$]² = a + $\frac{22a}{3}$

⇒ $\frac{25a^2}{9} = \frac{25a}{3}$

⇒ a = $\frac{9}{3}$

⇒ a = 3

∴ d =  $\frac{2}{3}$ × 3 = 2

Step 3:

We have the formula for the sum of first n terms as,

Sₙ =   [2a + (n-1)d]

substituting the values of a and d in the formula,

S₂₅ = $\frac{25}{2}$ [2×3 + (25 - 1)×2]  =  $\frac{25}{2}$ [6 +24×2]  =  $\frac{25}{2}$ × 54 = 25 × 27 = 675

 

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