Question

The 10th term and the 8th term of an A.P. are 25 and 41 respectively, then find the 38th term of that A.P.​

Answer 1

Answer:

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Explanation:

Let the first term be a and common difference be d then

a10=41⇒a+9d

a10 =73⇒a+17d

⇒a+9d+8d=73

⇒41+8d=73

⇒8d=73−41

⇒8d=32⇒d=4.

∴a+9d=41

⇒a+9×4=41

⇒a=5

Hence,

The 26th term =a+25d⇒5+25×4⇒105.

This is the required answer.

Answer 2

$\textsf{\large{\underline{Solution}:}}$

Let us assume that:

$\rm: \longmapsto a=First\ Term\ Of\ AP.$

$\rm: \longmapsto d=Common\ Difference.$

We know that:

$\rm: \longmapsto a_{n} = a + (n - 1)d$

Therefore:

$\rm: \longmapsto a_{10} = a + 9d = 25 - (i)$

$\rm: \longmapsto a_{8} = a + 7d =41 - (ii)$

Subtracting (ii) from (i), we get:

$\rm: \longmapsto (a + 9d) - (a + 7d)= 25 - 41$

$\rm: \longmapsto (a + 9d) - (a + 7d)= 25 - 41$

$\rm: \longmapsto 2d= - 16$

$\rm: \longmapsto d= -8$

Substituting the value of d in (i), we get:

$\rm: \longmapsto a + 9 \times ( - 8) = 25$

$\rm: \longmapsto a - 72= 25$

$\rm: \longmapsto a= 25 + 72$

$\rm: \longmapsto a= 97$

Now we have the values of a and d. Therefore, the 38th term will be:

$\rm: \longmapsto a_{38} = a + 37d$

$\rm = 97 + 37 \times ( - 8)$

$\rm = 97 - 37 \times8$

$\rm = 97 -296$

$\rm = - 199$

$\rm: \longmapsto a_{38} = - 199$

★ So, the 38th term of the A.P. is -199.

$\textsf{\large{\underline{Learn More}:}}$

Let a be the first term of an AP and d be the common difference. Then:

$\rm 1.\ \ a_{n}=a+(n-1)d$

$\rm 2.\ \ S_{n}=\dfrac{n}{2}[2a+(n-1)d]$