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

Question

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

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Madeline 1 month 2021-08-19T23:30:41+00:00 2 Answers 0 views 0

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    0
    2021-08-19T23:31:47+00:00

    Answer: answer is 59

    Step-by-step explanati

    0
    2021-08-19T23:32:32+00:00

    \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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