Question

Determine the ap whose fourth term is 18 and the difference of the ninth term fr the fifteenth term is 30

Answer 1

$\bold{GIVEN,}$

For an AP

$\huge \boxed {\mathsf{t_n = a + (n - 1)d}}$

Fourteen term = 18

${\mathsf{t_{4}= a + (4-1)d}}$

${\mathsf{18= a + 3d}}$----(1)

Difference between ninth term and Fifteenth term is 30

${\mathsf{t_{9} - t_{15}=30}}$

${\mathsf{t_{9} = a + 8d}}$

${\mathsf{t_{15} = a + 14d}}$

${\mathsf{a + 8d - (a + 14d )= 30}}$

${\mathsf{6d = 30}}$

${\mathsf{d = \frac{30}{6}}}$

$\huge \boxed {\mathsf{d = 5}}$

On placing value of d in eq (1)

${\mathsf{18= a + 3 \times 5 }}$

${\mathsf{18= a + 15 }}$

${\mathsf{18 - 15= a }}$

$\huge \boxed {\mathsf{a = 3}}$

FOR AP,

First term = a = 3

Second term = a+ d = 3+5=8

Third term = a + 2d = 3 + 10 = 13

So,

$\huge \boxed {\mathsf{A P = 3,8,13 ,18........}}$

Answer 2

$\bf\huge\textbf{\underline{\underline{According\: to\: the\: Question}}}$

4th term = a + 3d .......(Equation 1)

Now

(a + 8d) - (a + 14d) = 30

a + 8d - a - 14d = 30

6d = 30

$\sf{\implies d = \dfrac{30}{6}}$        

d = 5

$\bf\huge{\boxed{\bigstar{\sf\:{Put\:value\:of\:d\:in\:Equation\:(1)}}}}$

a + 3 × 5 = 18

a + 15 = 18

a = 18 - 15

a = 3

Therefore

a + d ⇒ 3 + 5 = 8

a + 2d ⇒ 3 + 10 = 13

$\bf\huge{\boxed{\bigstar{AP\:are\:3\:,8\:,13}}}$