Question

Find the sum of all terms of an A.P. if the common difference is -2, its first term
is 100 and the last term is - 10.​

Answer 1

$\\ \\ \large\underline{ \underline{ \sf{ \red{given:} }}}$

• First term ( a ) = 100

• Last term ( a{n} ) = -10

• Common difference ( d ) = -2

$\\ \\ \large\underline{ \underline{ \sf{ \red{to \: find:} }}}$

• Sum of all terms ( 100.......-10 )

$\\ \\ \large\underline{ \underline{ \sf{ \red{solution:} }}}$

We know ,

'n'th term , a{n} = a + ( n - 1 )d

Putting values , we get..

➻ -10 = 100 + ( n - 1 )-2

➻ -10 - 100 = -2n + 2

➻ -110 = -2n + 2

➻ 2n = 2+110

➻ 2n = 112

➻ n = 56

Hence , there are 56 terms.

We know ,

$\\ \\ \boxed{ \bf \: s_{n} = \frac{n}{2} (1st \: term + last \: term) }$

Here ,

• s{n} = sum of 'n' terms.

• n = 56.

• 1st term is 100.

• Last term is -10.

Putting values , we get..

$\sf \: s_{n} = \cancel\frac{56}{2} (100 + (- 10)) \\ \\ \sf \: s_{n} = 28(100 - 10) \\ \\ \sf \: s_{n} = 28(90) \\ \\ \boxed{ \bf{ \pink{ s_{n} = 2520 }}}$

Hence , sum is 2520.

Answer 2

To Find:-

• Find the sum of all the terms of an A.P and common difference.

Given:-

• The first term is 100 and last term is-10 and common difference is -2.

Solution:-

Here we know that

a = 100 ; l = -10 ; d = -2 ; s = ?

$\tt \implies { \: an \: = \: a \: + \: ( \: n \: - \: 1 \: ) \: d \: }$

$\tt \implies { \: -10 \: = \: 100 \: + \: ( \: n \: - \: 1 \: ) \: -2 \: }$

$\tt \implies { \: - 10 \: = \: 100 \: - \: 2n \: + \: 2 \: }$

$\tt \implies { \: - 10 \: = \: 102 \: - \: 2n \: }$

$\tt \implies { \: - 10 \: - \: 102 = \: - \: 2n \: }$

$\tt \implies { \: - 112 \: = - \: 2n \: }$

$\tt \implies { \: n \: = \: \frac{112}{2} \: }$

$\tt \implies { \: n \: = \: 58 \: }$

Sn = n/2 [ a + l ]

Sn = 58/2 [ 100 + ( -10 ) ]

Sn = 58/2 [ 90 ]

Sn = 58/2 × 90

Sn = 28 × 90

Sn = 2520