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If 7th term of an A.P is 1/9 and 9th term is 1/7 .
Find the 1st term of an A.P​

Answer 1

Answer:

answer is 1/21

Step-by-step explanation:

7th term=1/9

9th term =1/7

7th term = a+6d=1/9 -(1) -

9th term=a+8d=1/7 -(2) -

solving 1 and 2 equations

we get

a+6d=1/9

-a-8d=-1/7 ( here while solving sign of their terms changes)

=-2d=-2/63

=d=1/63

equate d=1/63 in equation one

a+6×1/63=1/9 ( here 6 and 63 are divisible by 3)

a+2×1/21=1/9

a+2/21=1/9

a=1/9-2/21

a=21-18/63

a=3/63

a=1/21

hope this answer helped you

Answer 2

$\sf\large{\underline{\blue{\underline{Question:-}}}}$

• If 7th term of an A.P is 1/9 and 9th term is 1/7 .
• Find the 1st term of an A.P

$\sf\large{\underline{\blue{\underline{Given:-}}}}$

• $\sf a_7=1/9$
• $\sf a_9= 1/7$

$\sf\large{\underline{\blue{\underline{To\:Find:-}}}}$

• $\sf a_{63}=?$

$\sf\large\maltese\:{\underline{\underline{Formula \:used :-}}}$

$\sf→{\fbox{\underline{\red{ a+(n-1)d}}}}$

$\sf→{\fbox{\underline{\red{a_n=a+(n-1)d}}}}$

$\sf\large{\underline{\blue{\underline{Solution:-}}}}$

$\sf→ a+(7-1)d=\frac{1}{9}\\\sf→ a+6d=\frac{1}{9}---equ(1)$

Similarly,

$\sf→ a+(9-1)d=\frac{1}{7}\\\sf→ a+8d=\frac{1}{7}---equ(2)$

Now,

Subtracting equ(1) from equ(2)

$\sf→ a+8d=\frac{1}{7}\\\sf→{\underline{_-a+_-6d=_-\frac{1}{9}}}\\\sf→ 2d=\frac{9-7}{63}\\\sf→ 2d=\frac{2}{63}\\\sf→ d={\fbox{\underline{\frac{1}{63}}}}$

Putting d= 1/63 in equation (1)

$\sf→ a+6d=\frac{1}{9}\\\sf→ a+6(\frac{1}{63})=\frac{1}{9}\\\sf→ a=\frac{-6}{6}+\frac{1}{9}\\\sf→ a=\frac{-6+7}{63}\\\sf→ a= {\fbox{\underline{\frac{1}{6}}}}$

Now,

For 63rd term.

$\sf→ a_n=a+(n-1)d\\\sf→ a_{63}=\frac{1}{63}+(63-1)\frac{1}{63} \\\sf→ a_{63}=\frac{1}{63}+\frac{62}{63}\\\sf→ a_{63}= \frac{1+62}{63}\\\sf→a_{63}=\frac{63}{63}\\\sf→{\fbox{\underline{ a_{63}= 1}}}$

$\sf\large{\underline{\blue{\underline{Hence:-}}}}$

$\sf→{\fbox{\red{\underline{\purple{ a_{63}= 1}}}}}$