Question

Let \(a_1,a_2,a_3,..\). be an increasing sequence of natural numbers, which are in an arithmetic progression with common difference d. Suppose \(a_1+a_2+a_3=27\) and \(a_1^2+a_2^2+a_3^2=275\). Then the value of \(a_1,d\) are 

• A. \(a_1=3,d=2\)
• B. \(a_1=4,d=5\)
• C. \(a_1=5,d=4\)
• D. \(a_1=2,d=3\)
• E. \(a_1=8,d=1\)

Answer 1

The Correct Option is C

Let the sequence be \( a_1, a_2, a_3 \) where \( a_2 = a_1 + d \) and \( a_3 = a_1 + 2d \). Given: \[ a_1 + a_2 + a_3 = 27 \implies 3a_1 + 3d = 27 \implies a_1 + d = 9 \quad (1) \] \[ a_1^2 + a_2^2 + a_3^2 = 275 \implies a_1^2 + (a_1 + d)^2 + (a_1 + 2d)^2 = 275 \] Expanding: \[ a_1^2 + (a_1^2 + 2a_1d + d^2) + (a_1^2 + 4a_1d + 4d^2) = 275 \] \[ 3a_1^2 + 6a_1d + 5d^2 = 275 \quad (2) \] From equation (1): \( a_1 = 9 - d \). Substitute into (2): \[ 3(9 - d)^2 + 6(9 - d)d + 5d^2 = 275 \] \[ 3(81 - 18d + d^2) + 54d - 6d^2 + 5d^2 = 275 \] \[ 243 - 54d + 3d^2 + 54d - 6d^2 + 5d^2 = 275 \] \[ 243 + 2d^2 = 275 \implies 2d^2 = 32 \implies d^2 = 16 \implies d = \pm 4 \] Since the sequence is increasing with natural numbers, \( d = 4 \). Then: \[ a_1 = 9 - d = 5 \] Final answer: \[ \boxed{a_1 = 5,\ d = 4} \]

Answer 2

Given that,
\(a_1,a_2,a_3…\) are in A.P
\(a_1+a_2+a_3=27\)      -------(1)
\(a_1^2+a_2^2+a_3^2=275\)       -------(2)
We know that,
\(a_2=a_1+d\)
\(a_3=a_1+2d\)
from the equation (1) 
\(a_1+(a_1+d)+(a_1+2d)=27\)
\(⇒a_1+d_1=3\)         -----(3) (Means \(a_2=9\))
from equation (2) 
\(a_1^2+a_2^2+a_3^2=275\)
\(⇒a_1^2+a_3^2=275-81\)
\(⇒a_1^2+a_3^2=194\)       -------(4)
So , Let us test from option to find the value of  \(a_1\) and \(d\)
From options mentioned, it can clearly be said that the value must be  4 and 5 with respect to equation (1)
let us check by taking \(a_1=5 \text{ and } d=4\)  that satisfies the equation 2 or not 
Hence , from equation (4)  we found that  the assumed values are respectively correct as LHS =RHS.

So, the correct option is (C): \(a_1=5,d=4\)