Question

10. The exterior angle and one in the opposite of one triangle are given here in
interior opposite angle and the third angle of each triangle.
(1) 120°, 45°
(ii) 95°, 35°
(iii) 58°, 20°
(iv) 130°, 60°
​

Answer 1

In the given AP, let first term = a and common difference = d

Then, Tn  =  a + (n-1) d

⇒ T8  = a + (8 – 1)d, T12  = a + (12 – 1)d

⇒ T8  = a + 7d, T12  = a + 11d

Now, T8  = 37 ⇒ a + 7d = 37  – – – (1)

T12  = 57 ⇒ a + 11d = 57  – – – (2)

Subtracting (1) from (2), we get

⇒ 4d = 20 ⇒ d = 5

Putting d = 5 in (1), we get

a + 7 x 5 = 37

⇒ a = 2

Tthus, a = 2, and d = 5

So the required AP is 2, 7, 12..