+4 marks−1 if incorrectNumericalPrevious-year question
Finding the Sum of an AP Using Determinant Conditions on Multiple Progressions
Let A_1, A_2, A_3 be the three A.P. with the same common difference d and having their first terms as A, A+1, A+2, respectively. Let a, b, c be the 7^th, 9^th, 17^th terms of A_1, A_2, A_3, respectively such that |[a 7 1; 2 b 17 1; c 17 1]|+70=0. If a=29, then the sum of first 20 terms of an AP whose first term is c-a-b and common difference is d/12, is equal to
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Hint 1 of 4
How can the terms a, b, and c be expressed in terms of the initial term A and common difference d?
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Correct answer
The sum of the first 20 terms of the specified arithmetic progression is 495.
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Solution
StepWorking
01Given
Three arithmetic progressions A1,A2,A3 have common difference d and first terms A,A+1,A+2 respectively. The terms a,b,c are the 7th,9th,17th terms of A1,A2,A3 respectively. Also, a2bc71717111+70=0 and a=29.
02Goal
Determine the sum of the first 20 terms of an AP with first term T1=c−a−b and common difference D=12d.
03Approach
Express a,b,c in terms of A and d: a=A+6d, b=(A+1)+8d=A+8d+1, c=(A+2)+16d=A+16d+2. Substitute into the determinant equation to find a linear relation between A and d. Combined with a=A+6d=29, solve for A and d. Finally, compute c−a−b and the sum S20.
04Execute
Substitute a,b,c into the determinant:
Δ=A+6d2(A+8d+1)A+16d+271717111=−70.
Apply row operations R2→R2−R3:
R2−R3=[2A+16d+2−(A+16d+2),17−17,1−1]=[A,0,0].
Expanding along R2:
Δ=−A71711=−A(7−17)=10A.
Given Δ+70=0⟹10A+70=0⟹A=−7.
Since a=A+6d=29, we get −7+6d=29⟹6d=36⟹d=6.
05Execute
Compute b and c:
b=A+8d+1=−7+8(6)+1=42.
c=A+16d+2=−7+16(6)+2=91.
First term of the required AP: T1=c−a−b=91−29−42=20.
Common difference of the required AP: D=12d=126=21.
Now compute the sum of the first 20 terms:
S20=220[2(20)+(20−1)(21)]=10[40+219]=10[299]=495.
✓Verify
Verify the determinant directly: a=29, 2b=84, c=91.
29849171717111=29−7917017101=−(−7)(7−17)=7(−10)=−70.
−70+70=0, which is consistent. The sum S20=495 is correct.
Hints that build this answer step by step
How can the terms a, b, and c be expressed in terms of the initial term A and common difference d?
a=A+6d, b=A+1+8d, c=A+2+16d
Evaluating the determinant Δ=a2bc71717111, what equation in a,b,c does Δ+70=0 yield?
Δ=10(2b−c)=−70⟹2b−c=−7
Using 2b−c=−7 and a=29, what are the values of A and d?
A=−7 and d=6
With A=−7 and d=6, what is the sum of the first 20 terms of the AP with first term A′=c−a−b and common difference d′=12d?
Why apply R2→R2−R3 rather than standard expansion?
Because rows 2 and 3 both have identical second and third column entries (17 and 1), subtracting them immediately yields two zeros, making expansion a one-line operation.
