Question

5th and 13th term of the AP are 5 and -3 resp. find its AP and also the 16th term

Answer 1

Answer:

-6

Step-by-step explanation:

T5=a+4d

a+4d=5

T13=a+12d

-3=a+12d

8d=-3-5=-8

d=(-1)

a+4×-1=5

a=9

T16=a+15d

=9+15×-1

=9-15

=(-6)

Answer 2

Step-by-step explanation:

$\bf{\underline{\underline\red{Given:-}}}$

• The 5th term of the AP is 5

• The 13th term of the AP is -3

$\bf{\underline{\underline\blue{To\:Find:-}}}$

• The 16th term of the AP

$\bf{\underline{\underline\green{Solution:-}}}$

As we know that:-

The nth term of an AP is given by the formula:-

$\boxed{ \rm \pink{\implies a_{n} = a + (n - 1)d }}$

So:-

5th term:-

$\rm \implies a_{5} = a + (5 - 1)d$

$\rm \implies a + 4d$

Given 5th term = 5

$\rm \implies a + 4d = 5......(i)$

13th term:-

$\rm \implies a_{13} = a + (13 - 1)d$

$\rm \implies a + 12d$

Given 13th term = -3

$\rm \implies a + 12d = -3......(ii)$

Multiplying equation (ii) with -1

$\rm \implies - 1(a + 12d = - 3)$

$\rm \implies - a - 12d = 3......(iii)$

Adding equation (i) and (iii)

$\rm \: \: \: \not a + 4d = 5$

$\rm - \not a - 12d = 3$

______________

$\rm \implies - 12d + 4d = 3 + 5$

$\rm \implies - 8d = 8$

$\rm \implies d = \dfrac{ \: \: \: 8}{ - 8}$

$\rm \implies d = - 1$

Substituting d = -1 in equation (i)

$\rm \implies a + 4 d = 5$

$\rm \implies a + 4 ( - 1) = 5$

$\rm \implies a - 4= 5$

$\rm \implies a = 5 + 4$

$\rm \implies a = 9$

We have :-

a = 9 and d = -1

Now 16th term:-

$\rm \implies a_{16}= a + (16 - 1)d$

$\rm \implies a + 15d$

$\rm \implies 9 + 15( - 1)$

$\rm \implies 9 - 15$

$\rm \implies - 6$

$\large\underline{ \boxed{ \rm \purple{ \therefore \: 16th \: term \: of \: the \: AP = - 6}}}$