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Find the sum of the following arithmetic progressions: (i) 50, 46, 42, … to 10 terms (ii) 1, 3, 5, 7, … to 12 terms
Arithmetic Progressions | CBSE | Class 10 | Exercise 9.6 | Maths | RD Sharma
Find the sum of the following arithmetic progressions: (i) 50, 46, 42, … to 10 terms (ii) 1, 3, 5, 7, … to 12 terms

In an A.P if first term is given by =a, common difference =d, and if no. of terms are n.

Therefore, sum of n terms of an A.P is given as:

{{S}_{n}}=n/2\left[ 2a+\left( n-1 \right) \right]

(i) Given A.P. =50, 46, 42, to 10 term.

First term (a) =50

Common difference (d) {{a}_{n}}-{{a}_{n-1}}=46-50=-4

Number of terms (n) =10

So, according to the formula,

{{S}_{10}}=10/2\left[ 2.50+\left( 10-1 \right)-4 \right]

=5{100-9.4}

=5{100-36}

=5\times 64

\therefore {{S}_{10}}=320

 

(ii) Given: A.P =1, 3, 5, 7, …..to 12 terms.

First term (a) =1

Common difference (d) ={{a}_{n}}-{{a}_{n-1}}=3-1=2

No. of term (n) =12

Therefore, sum of 12 terms ={{S}_{12}}=12/2\left[ 2.1+\left( 12-1 \right).2 \right]

=6\times \left\{ 2+22 \right\}

=6\times 24

\therefore {{S}_{12}}=144

i

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