Question

find the ap if the 6 th term of the ap is 19 and the 16 the term is 15 more than 11th term​

Answer 1

Step-by-step explanation:

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6th term of ap is 19

a+5d= 19...........(1)

again, 16th term of ap is 15 more than 11th term.

a+15d= 15+a+10d

5d= 15

d = 3

now putting the value d in equation 1

a+ 5×3= 19

a= 4

Ap is 4,7,10,13,16,19.....

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Answer 2

Answer:

$\bigstar{\bold{A.P\:is\:4,\:7,\:10,\:13.....}}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• 6th term ( a₆ ) = 19
• 16th term ( a₁₆) is 15 more than 11th term (a₁₁)

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The A.P

$\Large{\underline{\underline{\bf{Solution:}}}}$

➺ By given,

    a₆ = 19---------equation 4

    a₁₆ = a₁₁ + 15-------equation 3

➺ The nth term of an A.P is given by the formula

    $a_n=a_1+(n-1)d$

    $a_{16}=a_1+15d------equation\:1$

    $a_{11}=a_1+10d------equation\:2$

➺ Substitute equation 1 and 2 in equation 3

    a₁ + 15d = a₁ + 10d +15

➺ Cancelling a₁ on both sides,

    15d - 10d = 15

               5d = 15

                  d = 3

➺ Substitute the value of d in equation 4

    a₆ = 19

    a₁ + 5d = 19

    a₁ = 19 - 15

    a₁ = 4

➺ a₂ = a₁ + d = 4 + 3 = 7

➺ a₃ = a₂ + d = 7 + 3 = 10

➺ a₄ = a₃ + d = 10 + 3 = 13

➺ Hence A.P is 4, 7, 10, 13.........

$\boxed{\bold{A.P\:is\:4,\:7,\:10,\:13.....}}$

$\Large{\underline{\underline{\bf{Verification:}}}}$

➺ a₁₆ = a₁₁ + 15

    a₁ + 15 × 3 = a₁ + 10 × 3 + 15

    4 + 45 = 4 + 30 + 15

     49 = 49

Hence verified.

$\Large{\underline{\underline{\bf{Notes:}}}}$

➺ The common difference (d) of an A.P is the difference between its two consecutive terms.

   d = a₂ - a₁

➺ The nth term of an A.P is given by the formula,

    $a_n=a_1+(n-1)d$