Question

Find the sum of 0.6 1.7 2.8 to 100 terms​

Answer 1

Answer:

$\bf \color{red}Question :$

Find the sum of 0.6 1.7 2.8 to 100 terms

$\bf \purple{Solution: }$

$\bf \: Given,$

$\bf \: First \: term, a = 0.6$

$\bf \: Common \: difference, d = a_{2} - a_{1}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf= 1.7 -0.6$

$\bf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1.1$

$\bf \: Number \: of \: terms, n = 100$

Formula for sum of the nth term in the AP series is,

$\boxed{\color{blue}{\sf S _{n}= \frac{n}{2} \:[2a + (n − 1)d]}}$

$\bf \: S _{100} = \frac{100}{2} [2(0.6) + (100 – 1)1.1]$

$\: \: \: \: \: \: \: \: \: \: \: \bf= 50[1.2 + (99) \times (1.1)]$

$\: \: \: \: \: \: \: \: \: \: \: \bf= 50[1.2 + 108.9)$

$\: \: \: \: \: \: \: \: \: \: \: \bf= 50[110.1]$

$\: \: \: \: \: \: \: \: \: \: \: \bf= 5505$

$\bf \color{red}{Sum \: of \: first \: 100 \: terms = 5505}$

Answer 2

QUESTION:-

1. Find the sum of 0.6 1.7 2.8 to 100 terms.

ANSWER:-

$\tt \: →here \: a = 0.6$

$\tt \:→ d = 1.7 - 0.6 = 1.1 \: and,n = 100$

$\tt \: →we \: know \: that,$

$\tt \:→ {s}^{n} = \frac{n}{2} [ \: 2a + (n - 1)d]$

$\tt→∴ {s}^{100 = \frac{100}{2} } [ \: 2 \times 0.6 + (100 - 1) \times 1.1]$

$\tt \: →50 \times [2 \times 0.6 + (100 - 1) \times 1.1]$

$\tt→50 \times [1.2 + 108.9]$

$\tt→50 \times 110.1 = 5505.0$

$\tt \: ∴hence \: the \: sum \: of \: first \: 100 \: terms \: of \: the \: given \: A.P= 5505$

Hope it's help you