Question

The 19th term of an AP is equal to 3 times its 6th term. If its 9th term is 19, find the AP.​

Answer 1

$\Large\bold\red{\underline{\overline{\mid{\orange{solution:-}}\mid}}}$

t$\tiny{19}$=3t$\tiny{6}$

⇒3(a+(6−1)d)=a+(19−1)d

3(a+5d)=a+18d

3a+15d=a+18d

2a=3d

⇒d=$\dfrac{2}{3}$a

⇒t$\tiny{9}$=19

a+(9−1)d=19

a+8d=19

a+8×$\dfrac{2}{3}$a=19

3a+16a=19×3

19a=19×3

⇒a=3

⇒d=$\dfrac{2}{3}$a=$\dfrac{2}{3}$×3=2

∴ A.P is 3,5,7,9......

$\rm\underline\bold{hence\: solved \red{\huge{\checkmark}}}$

$\huge{\fcolorbox{a}{blue}{\fcolorbox{aqua}{aqua}{explore more}}}$

what is an arithmetic progression?

A:-a sequence of numbers in which each differs from the preceding one by a constant quantity (e.g. 1, 2, 3, 4, etc.; 9, 7, 5, 3, etc.).

Answer 2

Step-by-step explanation:

Given:-

• The 19th term of an AP is equal to 3 times its 6th term.

• The 9th term of the AP is 19.

To Find:-

• The AP.

Solution:-

Let " a " be the first term and " d " be the common difference.

Case 1:-

The 19th term is equal to three times the 6th term.

$\sf \implies a_{19} = 3(a_{6})$

$\sf \implies a + 18d = 3(a + 5d)$

$\sf \implies a + 18d = 3a + 15d$

$\sf \implies 3a - a + 15d - 18d = 0$

$\sf \implies 2a - 3d = 0......(i)$

Case 2:-

The 9th term of the AP is 19.

$\sf \implies a_{9} = 19$

$\sf \implies a + 8d = 19.....(ii)$

Multiply equation (ii) with 2:-

$\sf \implies 2(a + 8d = 19)$

$\sf \implies 2a + 16d = 38.....(iii)$

Equation (iii) - (i):-

$\sf \implies 2a + 16d - (2a - 3d) = 38$

$\implies \sf 2a + 16d - 2a + 3d = 38$

$\implies 19d = 38$

$\implies d = 2$

Substitute d = 2 in equation (ii)

$\sf \implies a + 8d = 19$

$\sf \implies a + 8(2) = 19$

$\sf \implies a = 19 - 16$

$\sf \implies a = 3$

First term = 3

Second term = a + d = 3 + 2 = 5

Third term = a + 2d = 3 + 4 = 7

Therefore:-

$\large\underline{\boxed{\therefore \textsf{\textbf{The \: AP = 3,5,7,9.....}}}}$