Question

put the numbers from 1 to 7 into the circle and no two consecutive numbers are allowed to be joined to each other, e.g.3 cannot be joined directly to 2 or 4

Answer 1

Answer:

Find the sum of the following arithmetic progressions:

(i) 50, 46, 42, … to 10 terms

(ii) 1, 3, 5, 7, … to 12 terms

(iii) 3, 9/2, 6, 15/2, … to 25 terms

(iv) 41, 36, 31, … to 12 terms

(v) a + b, a – b, a – 3b, … to 22 terms

(vi) (x – y)2, (x2 + y2), (x + y)2, to 22 tams

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 1

(viii) – 26, – 24, – 22, …. to 36 terms

Solution:

In an A.P if the first term = a, common difference = d, and if there are n terms.

Then, sum of n terms is given by:

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 2

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

First term (a) = 50

Common difference (d) = 46 – 50 = – 4

nth term (n) = 10

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 3

= 5{100 – 9.4}

= 5{100 – 36}

= S × 64

∴ S10 = 320

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

First term (a) = 1

Common difference (d) = 3 – 1 = 2

nth term (n) = 12

R D Sharma Solutions For Class 10 Maths Chapter 9 Arithemetic Progressions ex 9.6 - 4