​ Find the sums given below : 7 + 10 1/2 + 14 + …. + 84 ​

Question

Find the sums given below :

7 + 10\(\frac{1}{2}\) + 14 + …. + 84

Answer 1

Given A.P : 7 + 10\(\frac{1}{2}\) + 14 + …. + 84 

a = 7; d = a₂ – a₁ = 10\(\frac{1}{2}\) – 7 = 3\(\frac{1}{2}\) and the last term l = a_n = 84 

But, a_n = a + (n – 1) d

∴ 84 = 7 + (n – 1) 3\(\frac{1}{2}\) 

⇒ 84 – 7 = (n – 1) × \(\frac{7}{2}\) 

⇒ n – 1 = 77 × \(\frac{2}{7}\) = 22 

⇒ n = 22 + 1 = 23 

Now, S_n = \(\frac{n}{2}\) (a + l) where a = 7; l = 84 

S₂₃ = \(\frac{23}{2}\) (7 + 84) 

= \(\frac{23}{2}\) × 91 

= \(\frac{2093}{2}\)

= 1046\(\frac{1}{2}\)