Find the sums given below :7 + 10\(\frac{1}{2}\) + 14 + …. + 84

1.	Find the sums given below :7 + 10\(\frac{1}{2}\) + 14 + …. + 84
Answer» 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\(\frac{1}{2}\)

Answer»

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

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