Question

7 term of an A.P is 32 and 13 term of an A.P is 62 find A.P

Answer 1

Answer:

$\huge{\boxed{\boxed{\sf{a=2\:and\:d=5\:so\:A.P=2,7,12,17,22....}}}}$

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION-}}}}}$

□a7=32

□a13=62

○A.P=?

$a7 = a + 6d = 32 - - - - - (1) \\ a13 = a + 12d = 62 - - - - - (2) \\ subtracting \: (1) \: from \: (2) \\ = )12d - 6d = 62 - 32 \\ = )6d = 30 \\ = ) d = \frac{30}{6} \\ = )d = 5 \\ putting \: value \: of \: d \: in \: (2) \\ a + 12d = 62 \\ = )a + 12 \times 5 = 62 \\ = ) a = 62 - 60 \\ = )a = 2$

$\huge{\boxed{\boxed{\sf{a=2\:and\:d=5\:so\:A.P=2,7,12,17,22....}}}}$

Answer 2

$\bf Given:\ \sf a\ +\ 6d\ =\ 32\ and\ a\ +\ 12d\ =\ 62.\\\\\bf To\ find:\ \sf The\ A.P.\\\\\bf Answer:\ \\\\\sf a\ +\ 6d\ =\ 32\ \ \ \ \ \ \ \ \bigg(Equation\ 1 \bigg)\\a\ +\ 12\ d\ =\ 62\ \ \ \ \ \ \bigg(Equation\ 2 \bigg)\\\\\\On\ solving,\ we\ get\ d\ =\ 5.\ Substituting\ this\ value\ in\ equation\ 1,\\\\a\ +\ 6\ \times\ 5\ =\ 32\\\\a\ +\ 30\ =\ 32\\\\a\ =\ 32\ -\ 30\\\\a\ =\ 2$

$\sf Since\ we\ have\ the\ first\ term\ and\ the\ common\ difference,\ we\ are\ now\ able\ to\ find\ the\ A.P.\\\\Therefore,\ the\ A.P.\ is\ 2, 7, 12, 17, ...\ .$