Question

which term of an A.P 3,8,13,18..........,93​​

Answer 1

Solution:-

Given

$\rm \: first \: term \: (a) = 3$

$\rm \: common \: difference \: (d) = a_2 - a_1 = 8 - 3 = 5$

$\rm \: T_n = 93$

Using the formula

$\rm \: T_n = a + (n - 1)d$

Now put the value on formula

$\rm \: 93 = 3 + (n - 1) \times 5$

$\rm \: 9 3 = 3 + 5n - 5$

$\rm \: 93 - 3 + 5 = 5n$

$\rm \: 98 - 3 = 5n$

$\rm \: 95 = 5n$

$\rm \: n = \frac{95}{5}$

$\rm \: n = 19$

So number of term (n) = 19

Some information about AP

=> By an arithmetic progression of m terms, we mean a finite sequence of the form

General form of AP

a,a+d,a+2d,a+3d,...,a+(m−1)d.

=> The real number a is called the first term of the arithmetic progression, and the real number d is called the difference of the arithmetic progression.

Answer 2

${\bold{\blue{\boxed{\bf{Given}}}}}$

• first term = a = 3
• common diff. = d = 5
• nth term = 93

${\bold{\blue{\boxed{\bf{To\:find}}}}}$

• $\bold n = ???$

${\bold{\blue{\boxed{\bf{ Formulae\:used }}}}}$

$\bold a_n = a + (n-1)d$

${\bold{\blue{\boxed{\bf{ solution }}}}}$

$93 = 3 + (n -1)\times 5$

$93 - 3= ( n - 1 ) 5$

$90 = ( n - 1 ) 5$

$n - 1 = \dfrac{90}{5}$

⠀⠀⠀⠀

$n - 1 = 18$

⠀⠀

$n = 18 + 1$

⠀⠀⠀⠀

$n = 19$

${\bold{\blue{\boxed{\boxed{\bf{n = 19}}}}}}$