Question

the sum of first 6 terms of an ap is 42 the ratio of its 10th term to 13 term is 1 ratio 3 find the first and the 13th term of the AP​

Answer 1

Step-by-step explanation:

Sn = n/2 (2a + (n-1)d)

S6 = 6/2 (2a + (6-1)d) = 42

3( 2a + 5d) = 42

2a + 5d = 42/3

2a + 5d = 14 equ (1)

a10 : a13 : : 1 : 3

an = a + (n-1)d

a10 = a + (10 - 1)d = a + 9d

a13 = a + (13 - 1)d = a + 12d

a + 9d : a + 12d :: 1 : 3

$\frac{a + 9d}{a + 12d} = \frac{1}{3} \\ 3(a + 9d) = a + 12d \\ 3a + 27d = a + 12d \\ 3a + 27d - a - 12d = 0 \\ 2a + 15d = 0 \: \: \: \: \: \: \: equ \: (2)$

subtract equ (1) - (2)

2a + 5d = 14

2a + 15d = 0

-10d = 14

d = 14/-10

d = -1.4

substitute d = -1.4 in equ (1)

2a + 5d = 14

2a + 5(1.4) = 14

2a + 7 = 14

2a = 14 - 7

2a = 7

a = 7/2

a = 3.5

first term is a = 3.5

13th term

a13 = a + 12d

= 3.5 + 12(1.4)

= 3.5 + 16.8

= 20.3

hope you get your answer