Question

The 7th term of an A.P. is – 4 and its 13th term is – 16. Find the A.P.

Answer 1

Answer:

$\huge\underline{ \underline{ \red{your \: \: answer}}}$

Step-by-step explanation:

$\bf\red{ \mapsto \: a7 = - 4} \\ \\ \bf\pink{ \mapsto \:a + 6d = - 4 - - - - (1)} \\ \\ \bf\green{ \mapsto \:a13 = - 16} \\ \\ \bf\purple{ \mapsto \:a + 12 = - 16 - - - - - - (2)} \\ \\ \sf\blue{ (2) - (1) = = > }\\ \\ \bf\red{ \mapsto \:(a + 12d) - (a + 16d) = - 16 - ( - 4)} \\ \\ \bf\green{ \mapsto \: a + 12d - a - 16d = - 16 + 4} \\ \\ \bf\red{ \mapsto \: - 4d = - 12 }\\ \\ \bf\purple{ \mapsto \: d = \frac{ - 12}{ - 4} } \\ \\ \bf\orange{ \mapsto \:d = 3 } \\ \\ \bf\red{ value \: of \: d \: put \: in \: (1)} \\ \\ \bf\pink{ \mapsto \: a + 6d = - 4} \\ \\ \bf\blue{ \mapsto \: a = - 4 -( 6 \times 3)} \\ \\ \bf\green{ \mapsto \:a = - 4 - 18 } \\ \\ \bf\purple{\mapsto \:a = - 22 } \\ \\ \bf\red{ \underline{ \mapsto : \:a = - 22 \: and \: d = 3 }} \\ \\ \bf\orange{ Therefore \: the \: AP \: is\mapsto : </p><p>-22, -19, -16,-13......}$

$\large\boxed{ \fcolorbox{green}{blue}{hope \: it \: help \: you}}$

Answer 2

Answer:

$hey$

Step-by-step explanation:

a7 = a + 6d

-4 = a + 6d --------- eqn1

a13 = a + 12d

-16 = a + 12d ----------- eqn2

Subtract eqn1 from 2

We get

a + 12d = -16

- a + 6d = -4

__________

6d = - 12

__________

d = -12/6

d = -2

By putting the value of d in eqn1

a = -4 + { 6 (-2)}

a = -4 -12

a = -16

a1 = a = -16

a2 = a + d = -16 - 2 = -18

a3 = a + 2d = -16 -4 = -20

a4 = a + 3d = -16 -6 = -22

AP -> -16, -18, -20, -22, ............