Question

Find the sum of the first 14 terms of an ap in which 3rd term in 7 and 7th term is two more than thrice of its 3rd term

Answer 1

 S20 = ?                                        

S20 = n/2[2a + (n-1)d]                

a3 = a + 2d = 7    (given)            

a7 = 2 + 3(a3) = 2 + 3(7) = 2+21 = 23

a7 = a +6d = 23

a + 2d = 7 ....(1)

a + 6d = 23 .....(2)

by elimination method

(2) - (1)

                                 4d = 16

                                    d=4

substitute 'd' in (1)

                                     a + 8 = 7

                                          a = -1

S20 = 20/2 [2(-1) + (20 - 1) 4]

S20 = 10 [ -2 + 19 * 4]

S20 = 10 ( -2 + 76)

S20 = 10 ( 74)

S20 = 740 

Answer 2

Answer:

S14 = 350

Step-by-step explanation:

a3 = a + 2d = 7    (given)            

a7 = 2 + 3(a3) = 2 + 3(7) = 2+21 = 23

a7 = a +6d = 23

By Elimination Method

                          a + 2d = 7 --------(1)

                          a + 6d = 23 -------(2)

                                4d = 16

                                   d=4

substitute 'd' in (1)

                                    a + 8 = 7

                                         a = -1

S14 = 14/2 [2(-1) + (14 - 1) 4]

S14 = 7 [ -2 + 13* 4]

S14 = 7 ( -2 + 52)

S14 = 7 ( 50)

S14 = 350

∴The sum of first 14 terms is 350