Question

find the 32th term and nth term of AP : 101, 98 ,95, 92........
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Answer 1

Step-by-step explanation:

$\mathfrak{ \large \bold \red{ \underline{ \underline{given}}}} \\ \\ { \large \bold{101 \: \: \: 98 \: \: \: 95 \: \: \: \: 92 \: \: \: \: \: in \: ap }} \\ { \large \bold{we \: have \: to \: find \: 32 {}^{th} \: term \: of \: ap }} \\ \\ \\ { \large \bold \green{ \underline{step \: by \: step \: solution \: given \: below}}} \\ \\ { \large \bold{a = 101}} \\ { \large \bold{d = (98 - 101) = - 3}} \\ { \large \bold{n = 32}} \\ \\ { \large \bold \orange{ \underline{formula = a + (n - 1)d}}} \\ \\ { \large \bold{ = 101 + (32 - 1) \times ( - 3)}} \\{ \large \bold{ = 101 + (31) \times ( - 3)}} \\ { \large \bold{ = 101 - 93}} \\ { \large \bold{ = 8}} \\ \\ { \large \bold \blue{ \underline{answer = 8}}}$

Answer 2

Answer:32 th term=8,n th term=104-3n

Step-by-step explanation:first term (a1)=101,

common difference =-3,formula for n th term=a+(n-1)d

If 32th term= a+(n-1)d=101+(32-1)×-3

=101+(31)-3=101-93=8

If n th term=a+(n-1)×d=101+(n-1)×(-3)=101-3n+3=104-3n