Math, asked by haveeshjan2003, 9 months ago

Find the 14th term of te A.P whose 11th term is 38 and 16th term is 73

Answers

Answered by jungkooktae
1

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Answered by Anonymous
0

\boxed{\tt \dagger Given :- \dagger}

11th term of AP is 38 and,

16th term of AP is 73.

\boxed{ \tt \dagger To find :- \dagger}

The 31st term of AP = ?

\boxed{ \tt \dagger Solution :- \dagger}

Let first term of AP be a

Let first term of AP be aand common difference be d

Let first term of AP be aand common difference be dNow,

\tt \red{a_{11}=38a}

\tt\longrightarrow \green{a+10d=38\:.............(i)}

And,

 \blue{\tt\:a_{16}=73a}

 \tt\longrightarrow\pink{a+15d=73\:.............(ii)}

From eq (i) and eq (ii),

a + 10d = 38 ‿︵‿︵│

⠀ ⠀ ⠀ ⠀⠀⠀ ⠀ ⠀ ⠀⠀⠀ ⠀ |Subtracting

\boxed{+}a \boxed{+} 15d = \boxed{+}73 ‿︵‿︵│

-⠀ -⠀ ⠀ -

━━━━━━━━━━━━━━

-5d = -35

 \purple{ \tt⤇ d = \dfrac{-35}{-5} }

 \orange{ \tt⤇ d = 7}

Now,

Substitute the value of d in equation (i),

 \tt a + 10d = 38 \\ \tt⤇ a + 10 × 7 = 38 \\ \tt⤇ a + 70 = 38 \\ \tt⤇ a = 38 - 70 \\ \tt⤇ a = -32

Then,

 \gray{\tt\:a_{31}=a+30da }

\tt\longrightarrow\:a_{31}=-32+30\times{7} \\ \tt\longrightarrow\:a_{31}=-32+210 \\ \tt\longrightarrow\:a_{31}=178

Hence, the 31st term of an AP was \boxed{\sf\pink{178.}}

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