**Q) In an A.P., the sum of three consecutive terms is 24 and the sum of their squares is 194. Find the numbers.**

**Ans: **

**Step 1: **Let’s consider that the middle term is A, first term is A – D and third term is A + D, here D is the common difference.

By given 1st condition, sum of the three consecutive terms is 24.

Therefore, (A – D) + A + (A + D) = 24

∴ 3 A = 24

∴ A = 8

**Step 2:** By given 2nd condition, sum of the squares of three terms is 194

∴ (A – D)^{2} + A^{2} + (A + D)^{2} = 194

∴ (A^{2} + D^{2} – 2 A D) + A^{2} + (A^{2} + D^{2} – 2 A D) = 194

∴ 3 X^{2} + 2 D^{2} = 194

∴ 3 (8)^{2} + 2 D^{2} = 194

∴ 2 D^{2} = 194 – 3 x 64

∴ 2 D^{2} = 194 – 192 = 2

∴ D^{2} = 1

∴ D = 1

**Step 3:** Since we considered the three consecutive terms as (A – D), A, (A + D) and calculated A = 8 and D = 1

Now, the three terms are (8 – 1), 8, (8 + 1) or 7, 8 ,9

**Therefore, the three terms are 7, 8 ,9.**

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