Question

in an arithmetic progression the sum of the 3rd and the 15th term is "-34" also the sum of 6th and 20th term of "-58". find. d​

Answer 1

Answer:

hey here is your solution

pls mark it as brainliest

Step-by-step explanation:

$to \: find = \\ common \: difference \: (d) \\ \\ so \: for \: a \: certain \: A.P \\ we \: know \: that \\ \\ tn = a + (n - 1)d \\ \\ hence \: then \\ 3rd \: term = a + (3 - 1)d \\ = a + 2d \\ \\ 15th \: term = a + (15 - 1)d \\ = a + 14d \\ \\ 6th \: term = a + (6 - 1)d \\ = a + 5d \\ \\ 20th \: term = a + (20 - 1)d \\ = a + 19d$

$according \: to \: first \: condition \\ a + 2d + a + 14d = - 34 \\ \\ 2a + 16d = - 34 \\ \\ a + 8d = - 17 \: \: \: \: \: \: (1) \\ \\ according \: to \: second \: condition \\ a + 5d + a + 19d = - 58 \\ \\ 2a + 24d = - 58 \\ \\ a + 12d = - 29 \: \: \: \: \: \: \: \: \: (2)$

$applying \: now \\ (1) - (2) \\ we \: obtain \\ \\ (8d - 12d) = - 17 - ( - 29) \\ \\ - 4d = - 17 + 29 \\ \\ - 4d = 12 \\ \\ d = - 3$