Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept 🎬 GP → AP Concept in 59 Sec Kabhi question bole: 👉 “Numbers GP mein hain, par kuch subtract/add karne ke baad AP ban gaye” Yeh coincidence nahi hota — yeh pure logic + standard form ka game hai 🔥 🖼️ Concept Image ✍️ Short Explanation Is type ke questions mein students ki sabse badi galti hoti hai: ❌ numbers ko direct manipulate karna JEE ka smart approach hai: 👉 GP ko standard form mein likho 👉 AP ki difference condition lagao 👉 a aur r cleanly nikal jaate hain 🔹 Step 1 — Write GP in Standard Form (CONTROL STEP 🔥)** Agar x 1 ,   x 2 ,   x 3 ,   x 4 x_1,\ x_2,\ x_3,\ x_4 GP mein hain, toh hamesha likho: x 1 = a , x 2 = a r , x 3 = a r 2 , x 4 = a r 3 x_1=a,\quad x_2=ar,\quad x_3=ar^2,\quad x_4=ar^3 📌 Bas yahin se poora question control mein aa jaata hai. 🔹 Step 2 — Apply the Given Operation Question mein jo operation diya ho (usually subtraction/addition), apply karo: ( x 1 − k 1 ) ,   ( x 2 − k 2 ) ,   ...

❓ 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 ...