Question

The 10th term of the AP 1,3/2,2,5/2 ...... is *

Answer 1

$\large \orange{answer} \\ \\ \large \implies{t \: 10 = \frac{11}{2} } \\ \\ \large \implies{ \underline{given}} \\ \\ \large \implies{ap \: = \: 1 + \frac{3}{2} + 2 + \frac{5}{2} ...... \: n \: terms } \\ \\ \large \implies \underline{find \: \: 10 \: terms} \\ \\ \large \implies \underline{ solutions} \\ \\ \large \implies{first \: \: term \: \: a = 1} \\ \\ \large \implies{common \: \: difference \: \: d \: = \frac{3}{2} - 1 = \frac{1}{2} } \\ \\ \large \implies{t \: 10 \: terms \: = a + 9d} \\ \\ \large \implies{1 + 9 \times \frac{1}{2} } \\ \\ \large \implies \therefore{t \: 10 = \frac{11}{2} }$

Answer 2

QUESTION:

The 10th term of the AP 1,3/2,2,5/2 ...... is *

FORMULA USED :

$</u><u>\</u><u>h</u><u>u</u><u>g</u><u>e</u><u>\</u><u>o</u><u>r</u><u>a</u><u>n</u><u>g</u><u>e</u><u> </u><u>{</u><u>nth \: term = a + (n - 1)d</u><u>}</u><u>$

where;

a = first term

d = common difference

Now come to main question;

AP :

$\red {1... \frac{3}{2} ....2.... \frac{5}{2}}$

here;

first term = 1

common difference = 3/2 - 1

=3-2 /2

= 1/2

Using the formula;

$\blue {t(10) = 1 + (10 - 1) \times \frac{1}{2} </p><p>}$

$t(10) = 1 + 9 \times \frac{1}{2} \\ t(10) = 1 + \frac{9}{2} \\ t(10) = \frac{2 + 9}{2} \\ t(10) = \frac{11}{2}$

$\huge\purple {10th \: term = \frac{11}{2} }$