Question

the 7th term of an AP is -1 and its 16th term is 17. The nth term of the ap is? ​

Answer 1

$hey \: mate \: i \: think \: i \: can \: solve \: it$

T7 = -1 eq 1

t16 =17 eq2

.in eq 1

we can write t7 that is seventh term in another way using a formula

that is tn = a + (n-1)d

so a +(7-1)d=-1

a +6d=-1 eq 3

in eq 2

t16 = 17

a+15d =17 eq4

now subtract eq 3 and eq4

a+ 6d = -1

-a-15d=-17

__________

-9d =-18

__________

so d = 2

them substitute d = 2 in in any equation

but I am substituting it in eq 3

a + 6(2) = -1

a = -13

now

the question is to find tn

so

tn = a+(n-1)d

tn= -13 + (n-1)2

tn = -13 +2n-2

tn = 2n-15

_________

hope this helps you

$thankuuuuuuu$

Answer 2

Answer:

7th term of AP = -1

⇒ a + (n-1)d = -1

a + (7-1)d = -1

a + 6d = -1

a = -1 - 6d — (i)

16th term of AP = 17

⇒ a + (n-1)d = 17

a + (16-1)d = 17

a + 15d = 17

a = 17 - 15d — (ii)

Now we compare both the equations:-

Since we know that both the equatios stand for the value of ‘a’, we can say that they are equal to each other.

-1 - 6d = 17 - 15d

15d - 6d = 17 + 1

9d = 18

d = 2

Now we find the value of ‘a’ by substituting the value of ‘d’ in equation (i):-

a = -1 - 6d

a = -1 - 6(2)

a = -1 - 12

a = -13

Now we find the value of ‘n’th term of AP:-

Formula = a + (n-1)d

⇒ -13 + (n-1)2

-13 + 2n - 2

-15 + 2n OR 2n - 15

Nth term of AP = 2n - 15

Hope it helps

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