Question

Find the A.P. whose 7th and 12th terms are 34 and 64 respectively. Also find its 10th term

Answer 1

$we \: know \: that \: in \: a \: A.P \: the \: {n}^{th} \: term \: is \: given \: by \: \\ a_{n} = a + (n - 1)d \: \\ where \: a \: = first \: term \: and \: d \: = \: common \: difference \: \\ {7}^{th} \: term \: = 34 \\ i.e \: \: a_{7} = a + 6d = 34.......(1) \\ \\ {12}^{th} \: term \: = 64 \\i.e \: a_{12} = a + 11d = 64......(2) \\ subtract \: equation \: (1) \: from \: equation \: (2) \: we \: get.. \\ 5d = 30 \\ = > d = 6 \\ now \: plug \: in \: the \: value \: of \: d \: in \: equation \: (1) \\ a + 36 = 34 \\ = > a = - 2$

So the required A.P will be
a , (a+d) , (a+2d)....... {a+(n-1)d}

plug-in the values of 'a' and 'd' we get
A.P. = -2 , 4 , 10 , 16.........

$now \: we \: have \: to \: find \: the {10}^{th} \: term. \\ = a _{10} = a + 9d \\ plug \: in \: the \: values \: of \: \: a \: and \: d \: \\ = > a _{10} = - 2 + 54 = 52 \\ hence \: {10}^{th} \: term \: of \: the \: A.P \: is \: 52$