Let aₙ be the nᵗʰ term of an A.P. If Sₙ = a₁ + a₂ + a₃ + ⋯ + aₙ = 700, a₆ = 7 and S₇ = 7, then aₙ is equal to:
❓ Question Let a n a_n a n be the n th n^{\text{th}} term of an A.P. If S n = a 1 + a 2 + ⋯ + a n = 700 , a 6 = 7 S_n=a_1+a_2+\cdots+a_n=700,\; a_6=7 and S 7 = 7 S_7=7 , then find a n a_n . 🖼️ Question Image ✍️ Short Solution Let the A.P. have first term a 1 a_1 and common difference d d . Then: a 6 = a 1 + 5 d = 7. Sum of first 7 terms: S 7 = 7 2 ( 2 a 1 + 6 d ) = 7. S_7 = \frac{7}{2}\big(2a_1 + 6d\big) = 7. Divide both sides by 7: 1 2 ( 2 a 1 + 6 d ) = 1 ⇒ a 1 + 3 d = 1. \frac{1}{2}\big(2a_1 + 6d\big) = 1 \quad\Rightarrow\quad a_1 + 3d = 1. Now subtract the two linear equations: ( a 1 + 5 d ) − ( a 1 + 3 d ) = 7 − 1 ⇒ 2 d = 6 ⇒ d = 3. (a_1+5d) - (a_1+3d) = 7 - 1 \;\Rightarrow\; 2d = 6 \;\Rightarrow\; d = 3. Then a 1 + 3 d = 1 ⇒ a 1 + 9 = 1 ⇒ a 1 = − 8. a_1 + 3d = 1 \Rightarrow a_1 + 9 = 1 \Rightarrow a_1 = -8. So the A.P. is − 8 , − 5 , − 2 , 1 , 4 , 7 , 10 , … Next, use the condition S n = 700 S_n = 700 . Sum ...