Question

show that a1,a2,....,an...form an APwhere an is defined as below : [a] an = 3 + 4n [b] an = 9 - 5n this question is from ncert mathematics book of class 10 exercise 5.3 [ques.no.10]

Answer 1

Answer:

an=3+4n

a1= 3 + 4(1) = 7

a2 = 3+ 4(2) = 11

a3 = 3+4(3) = 15

..d=a3-a1 = 11 - 7=4

Here a = 7, d = 4 and n = 15

By using sn =n/2 {2a + (n – 1)d) we have,

S15 =15[2 x 7+ (15 - 1)4]

=15/2 (14+56)

= 15/2x 70 = 525

(ii) an = 9 - 5n given

a1= 9 -5(1) = 9 5 = 4

a2=9-5(2) = 9 -10 = -1

a3= 9 -5(3) = 9 - 15 = -6

d = a2– a1 = -6 - (-1)= -5

Here a = 4, d= -5 and n = 15

By using Sn =n/2[2a + (n − 1)d] we have,

S15=15/2[2 x 4 + (15 1)(-5)]

=15/2(8 – 70)

=15/2x (-62) = -465

Step-by-step explanation:

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