Question

10th term of an ap is 30 and 7th term is 8 find 17th term?​

Answer 1

$\large\bf\underline \blue {To \: \mathscr{f}ind:-}$

• we need to find the 17th term

$\huge\bf\underline \purple{ \mathcal{S}olution:-}$

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

• 10th term of AP = 30
• 7th term of AP = 8

we know that,

$\large \bigstar \dag \red{\bf \: a_n = a + (n -1 )d}$

So,

⇝10th term = a + 9d = 30.....1)

⇝7th term = a + 6d = 8 ......2)

• From equation 1) and 2)

⠀⠀⠀⠀⠀a + 9d = 30

⠀⠀⠀⠀⠀a + 6d = 8

⠀⠀⠀⠀⠀-- ⠀⠀--⠀--⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀⠀3d = 22

⠀⠀⠀⠀⠀⠀⠀⠀d = 22/3

• putting value of d in 1)

↛ a + 9d = 30

↛ a + 9 × 22/3 = 30

↛ a + 66 = 30

↛ a = 30 - 66

• → a = -36

Now,

• 17th term of AP = a + 16d

↛ a17 = -36 + 16 × 22/3

↛ a17 = -36 + 352/3

↛ a17 = -108 + 352/3

↛ a17 = 244/3

Hence,

• 17th term of AP is 244/3

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Answer 2

$\it\red{\underline{\underline{Answer:}}}$

$\sf{The \ 17^{th} \ term \ of \ AP \ is \ \dfrac{244}{3}.}$

$\it\orange{Given:}$

$\sf{In \ an \ AP,}$

$\sf{\longmapsto{t_{10}=30}}$

$\sf{\longmapsto{t_{7}=8}}$

$\it\pink{To \ find:}$

$\sf{17^{th} \ term \ of \ the \ AP.}$

$\it\green{\underline{\underline{Solution:}}}$

$\boxed{\sf{t_{n}=a+(n-1)d}}$

$\sf{According \ to \ the \ first \ condition. }$

$\sf{t_{10}=a+9d}$

$\sf{\therefore{a+9d=30...(1)}}$

$\sf{According \ to \ the \ second \ condition. }$

$\sf{t_{7}=a+6d}$

$\sf{\therefore{a+6d=8...(2)}}$

$\sf{Subtract \ equation(2) \ from \ equation(1), \ we \ get}$

$\sf{a+9d=30}$

$\sf{-}$

$\sf{a+6d=8}$

__________________

$\sf{3d=22}$

$\boxed{\sf{\therefore{d=\dfrac{22}{3}}}}$

$\sf{Substitute \ d=\dfrac{22}{3} \ in \ equation(2), \ we \ get}$

$\sf{\leadsto{a+6\times\dfrac{22}{3}=8}}$

$\sf{\leadsto{a+44=8}}$

$\sf{\leadsto{a=8-44}}$

$\boxed{\sf{\leadsto{a=-36}}}$

$\sf{Now,}$

$\sf{a=-36 \ and \ d=\dfrac{22}{3}}$

$\sf{But, \ t_{n}=a+(n-1)d}$

$\sf{\therefore{t_{17}=a+16d}}$

$\sf{\therefore{t_{17}=-36+16\times\dfrac{22}{3}}}$

$\sf{\therefore{t_{17}=-36+\dfrac{352}{3}}}$

$\sf{\therefore{t_{17}=\dfrac{-108+352}{3}}}$

$\sf{\therefore{t_{17}=\dfrac{244}{3}}}$

$\sf\purple{\tt{\therefore{The \ 17^{th} \ term \ of \ AP \ is \ \dfrac{244}{3}.}}}$