Question

for an AP t7 =10 t13 =34 find t10​

Answer 1

Answer:

22 is tge t10

d = 4

a = -14 so a+9d=a10

-14+9 ✖ 4 =22

Answer 2

$\large\color{magenta}\qquad \qquad \underline{ \pmb{ \mathbb{ \maltese \: \: GIVEN \: \: \maltese}}}$

$\large\mathfrak{ \text{F}or \: \: an \: \: \bf A.P. } \\ \\ \bf t_7 = 10 \\ \\ \bf t_{13} = 34$

$\large\color{red}\qquad \qquad \underline{ \pmb{ \mathbb{ \maltese \: \: TO \: \: FIND \: \: \maltese}}}$

$\large\bf t_{10} \\ \bf i.e. \\ \bf 10th \: term \: of \: the \: AP$

$\color{cyan}\qquad \qquad \underline{ \pmb{ \mathbb{ \maltese \: \: KEY \:\:CONCEPT\: \: \maltese}}}$

$\mathfrak{nth \: \: term \: \: of \: \: an \: \: \text{ AP }\: is \: } \\ \\ \bf t_n = a + (n - 1)d$

$\large \mathcal{WHERE}$

$\mathfrak{a = first \: \: term \: \: of \: \: the \: \: \bf AP} \\ \\ \mathfrak{ \text{d} =common \: \: difference \: \: of \: \: \bf AP }$

$\large\color{purple}\qquad \qquad \underline{ \pmb{ \mathbb{ \maltese \: \: SOLUTION\: \: \maltese}}}$

$\bf \Large As\\\\\sf t_7 = 10 \\ \implies \bf a + 6d = 10 \: \: \: \: ...(i)\\ \\ \sf t_{13} = 34 \\ \implies \bf a + 12d = 34 \: \: \: \: ...(ii)$

Now,

$\bf (ii) - (i) \implies \\ \\ \sf 6d = 24 \\ \\ \implies \green{ \boxed{ \boxed{ \bf d = 4}}}$

Putting value of d in (i)

$\implies \green{\boxed{ \boxed{ \bf a = - 14}}}$

At Last

$\bf t_{10} = a + 9d \\ \\ = - 14 + 9(4) \\ \\ = - 14 + 36 \\ \\ \implies \huge \pink{ \boxed{ \boxed{ \bf t_{10} {= 22}}}}$