Question

four no are in ap with d>0 if their sum is 20 and sum of their squares find middle term​

Answer 1

Answer:

Let the numbers be a-2d, a-d, a and a+d. Their sum is

a-2d+ a-d+ a + a+d = 4a-2d = 20, or

2a - d = 10 …(1)

(a-2d)^2+ (a-d)^2+ a^2 + (a+d)^2 = 180, or

a^2–4ad+4d^2+a^2–2ad+d^2+a^2+a^2+2ad+d^2 = 180, or

a^2–4ad+4d^2+a^2+d^2+a^2+a^2+d^2 = 180, or

4a^2–4ad+6d^2 = 180, or

2a^2–2ad+3d^2 = 90 …(2)

From (1) d = 2a -10. Put that in (2) to get

2a^2 -2a(2a–10) + 3(2a–10)^2 = 90, or

2a^2+20a-4a^2+3(4a^2-40a+100) = 90, or

2a^2+20a-4a^2+300–120a+12a^2 = 90, or

10a^2–100a+210 = 0, or

a^2–10a+21=0

(a-3)(a-7) = 0

Hence a = 3 or 7 and the corresponding value of d from (1)

d=2a - 10 is -4 or 4.

So the 4 terms of the AP are 3-[2*(-4)] = 11, 7, 3 and -1, or

-1, 3, 7 and 11.

Check: 11+7+3–1 = 20. Correct.

1^2+3^2+7^2+11^2 = 1+9+49+121 = 180. Correct.

Answer 2

$\huge{\mathcal{\underline{\green{QuEsTiOn}}}}$Prove$\sqrt{ \frac{1 + \sin( \alpha ) }{1 - \sin( \alpha ) } } = \sec\alpha + \tan \alpha$

see attachment

hope it helps you ✌️