Question

The sum of first six terms of an arithmetic progression is \(345\) and the difference between the first term and the sixth term is \(55\). Its first term and common difference are respectively

• A. \(85,-11\)
• B. \(85,11\)
• C. \(-85,11\)
• D. \(-85,-11\)

Hint:

In A.P. problems, first form equations using sum formula and nth term formula.

Answer 1

The Correct Option is A

Concept: For an A.P., \[ S_n=\frac{n}{2}[2a+(n-1)d] \] and \[ T_n=a+(n-1)d \] where \(a\) is first term and \(d\) is common difference.

Step 1: Use sum of first 6 terms. Given: \[ S_6=345 \] So, \[ \frac{6}{2}[2a+5d]=345 \] \[ 3(2a+5d)=345 \] \[ 2a+5d=115 \qquad ...(1) \]

Step 2: Use difference between first and sixth terms. First term: \[ T_1=a \] Sixth term: \[ T_6=a+5d \] Given difference: \[ a-(a+5d)=55 \] \[ -5d=55 \] \[ d=-11 \]

Step 3: Find first term. Substitute in (1): \[ 2a+5(-11)=115 \] \[ 2a-55=115 \] \[ 2a=170 \] \[ a=85 \] Thus, \[ a=85,\quad d=-11 \] \[ \boxed{(85,-11)} \]