Question

141.
If the third and the tenth term of an arithmetic progression are 14 and 56, respectively,
then the arithmetic mean of the first 15 terms of the arithmetic progression is
(A)
(C)
40
80
(B) 44
(D) 34​

Answer 1

Answer:

44

Step-by-step explanation:

Attached as photo

Hope it helps you

Answer 2

Answer:

arithmetic mean of first 15th term is  = 90

Step-by-step explanation:

Given :- t3 = 14 , t10 = 56

To find :- arithmetic means of first 15th terms .

Solution :-

Step 1) Formula we are going to use are :-

a) tn = a+(n-1)d , where a = first term , n = no. of term , d = common difference .

b) arithmetic mean = 2a + l , where , l = last term .

Step 3)

t3 = a + (3-1)d

14 = a + 2d ----(1)

t10 = a + (10-1)d

56 = a + 9d ----(2)

solving eqn. (1) and (2) simultaneously ,

we get ,

a = 2 , d = 6 .

therefore , t15 = a + (15-1) d

                   t15 = 2+ 14*6

                   t15 = 86 .

Step 3)  Arithmetic mean of first 15th terms is = 2*a+l

                                                                   M = 2*2 + 86

therefore , mean = 90

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