Question

for an ap,if t7=32,t13=62,find t30​

Answer 1

Answer:

$let \: the \: first \: term \: of \: ap \: is \: a \: and \: common \: difference \: is \: d \\ now \: t7 = a + (7 - 1)d \: = 32 - - - - - (1) \\ t13 = a + (13 - 1)d = 62 - - - - - - (2) \\ now \: by \: (2) - (1) \: \\6d = 30 \\ or \: d = 5 \\ puting \: the \: value \: of \: d \: in \: eqation\: (1) \\ a = 2 \\ so \: t30 \: = a + 29d = 2 + 29 \times 5 = 147$

i hope you will understand....

Answer 2

GIVEN :

$\sf {t}_{7} = a + 6d= 32 \\ \\ \sf{t}_{13} = a + 12d = 62$

SOLUTION :

$\sf By \: subtracting \: {t}_{7} \: from \: {t}_{13} \\ \\ \sf a + 12d - (a + 6d) = 62 - 32 \\ \\ \sf \cancel{ a }+ 12d \: \cancel{ - a }- 6d = 30 \\ \\ \sf 6d = 30 \\ \\ \large \boxed{ \sf \blue{d = 5}} \\ \\ \sf put \: value \: of \: d \: in \: {t}_{7} \\ \\ \sf a + 6 \times 5 = 32 \\ \\ \sf a = 32 - 30 \\ \\ \large \boxed{ \sf \green{ a = 2}}$

$\sf {t}_{30} = a + 29d \\ \\ \sf {t}_{30} = 2 + 29 \times 5 \\ \\ \large \boxed{ \sf \orange{{t}_{30} = 147}}$