Question

a 19th term of an ap is equal to the three times ots 6th term .If its 9th term is 19.find the ap.​

Answer 1

$\bf \to \: \green{{ \underline{given \div }}}$

$\sf \to \: 19th \: term \: of \: an \: ap \: is \: equal \: to \: three \: times \: to \: 6th \: term \\ \\ \sf \to \: if \: its \: 9th \: term \: = 19$

$\bf \to \: \pink{{ \underline{to \: find \: the \: ap \div }}}$

$\bf \to \: \orange{{ \underline{step \: - \: by \: - \: step \: - explanation}}}$

$\sf \to \: 19th \: term \: = 3(6th \: term) \\ \\ \sf \to \: \: { \underline{formula \: of \: nth \: term \: of \: an \: ap}} \\ \\ \sf \to \: a_{n} \: = a \: + \: (n - 1)d \\ \\ \sf \to \: a \: + 18d \: = 3(a \: + 5d) \\ \\ \sf \to \: a \: + 18d \: = 3a \: + 15d \\ \\ \sf \to \: 3d \: = 2a \:....(1) \\ \\ \sf \to \: 9th \: term \: = 19 \\ \\ \sf \to \: a \: + 8d = 19.....(2)$

$\\ \\ \sf \to \: from \: equation \: (1) \: we \: get \\ \\ \sf \to \: a \: = \frac{3d}{2} ......(3) \\ \\ \sf \to \: put \:equation \: (3) \: in \: (2) \\ \\ \sf \to \: \frac{3d}{2} + 8d \: = 19 \\ \\ \sf \to \: 3d \: + 16d \: = 38 \\ \\ \sf \to \: 19d \: = 38 \\ \\ \sf \to \: d \: = 2 \\ \\ \sf \to \: put \: the \: value \: of \: d \: = 2 \: in \: equation \: (3) \\ \\ \sf \to \: a \: = \frac{3 \times 2}{2} = 3 \\ \\ \sf \to \: first \: term \: = 3 \\ \\ \sf \to \: common \: difference \: = 2$

$\\ \\ \sf \to \: first \: term \: = a \: = 3 \\ \\ \sf \to \: second \: term \: = a \: + d \: = 3 + 2 = 5 \\ \\ \sf \to \: third \: term \: = a \: + 2d \: = 3 + 2 \times 2 = 7 \\ \\ \sf \to \: fourth \: term \: = a + 3d \: = 3 + 3 \times 2 = 9 \\ \\ \sf \to \: \green{{ \underline{sequence \: is \: = 3 \:, 5 \: ,7 \:, 9 ,\: ......}}}$

Answer 2

Step-by-step explanation:

t

19

=3t

6

⇒3(a+(6−1)d)=a+(19−1)d

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

3a+15d=a+18d

2a=3d

⇒d=

3

2

a

⇒t

9

=19

a+(9−1)d=19

a+8d=19

a+8×

3

2

a=19

3a+16a=19×3

19a=19×3

⇒a=3

⇒d=

3

2

a=

3

2

×3=2

∴ A.P is 3,5,7,9......

Hence, solved.