In an AP,sum of first 8 terms is 136 and that of 15 terms is 465. Find the sum of 25 terms.​

Question

In an AP,sum of first 8 terms is 136 and that of 15 terms is 465. Find the sum of 25 terms.​

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Ivy 1 month 2021-10-27T02:42:11+00:00 2 Answers 0 views 0

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    0
    2021-10-27T02:43:45+00:00

    *۰۪۫G۪۫۰۰۪۫i۪۫۰۰۪۫v۪۫۰۰۪۫e۪۫۰۰۪۫n۪۫۰:

    • Sum of first 8 terms(S8)=136
    • Sum of 15 terms (S15) =465

    * ۰۪۫T۪۫۰۰۪۫o۪۫۰ ۰۪۫F۪۫۰۰۪۫i۪۫۰۰۪۫n۪۫۰۰۪۫d۪۫۰:

    • Sum of 25 terms(S25) =?

    * ۰۪۫S۪۫۰۰۪۫o۪۫۰۰۪۫l۪۫۰۰۪۫u۪۫۰۰۪۫t۪۫۰۰۪۫i۪۫۰۰۪۫o۪۫۰۰۪۫n۪۫۰

    Formula for sum of an A. P :

    _____________

    Sn= n/2[2a+(n-1) d]

    _____________

    Now to find the sum of first 8 terms,

    By Using formula,

    S8=8/2[2a+(81) d]

    136=4[2a+7d]

    136/4=2a+7d

    34=2a+7d

    2a+7d=34...................... (1)

    Now to find the Sum of 15 terms,

    By using formula,

    S15=15/2[2a+(151) d]

    465=15/2[2a+14d]

    465×2/15=2a+14d

    31×2=2a+14d

    62=2a+14d

    62=2(a+7d)

    62/2=a+7d

    31=a+7d

    a+7d=31........................(2)

    Subtracting equation (2) From (1) we get,

    a=3

    Now Substituting the value of a in (2)

    3+7d=31

    7d=313

    7d=28

    d=28/7

    d=4

    Now, Let’s find the Sum of 25 terms.

    By using formula,

    S25=25/2[2×3+(251) 4

    =25/2[6+24×4]

    =25/2[6+96]

    =25/2×102

    =25×51

    =1275

    Therefore, The sum of 25 terms is 1275.

    0
    2021-10-27T02:43:56+00:00

    Solution :

    \bigstar Using formula of the sum of an A.P;

    \boxed{\bf{S_n=\frac{n}{2} \bigg[2a+(n-1)d\bigg]}}}}

    • a is the first term
    • d is the common difference
    • n is the term of an A.P.

    A/q

    \longrightarrow\sf{136=\cancel{\dfrac{8}{2}} \bigg[2a+(8-1)d\bigg]}\\\\\longrightarrow\sf{136=4[2a+7d]}\\\\\longrightarrow\sf{\cancel{136/4}=2a+7d}\\\\\longrightarrow\sf{34=2a+7d........................(1)}

    &

    \longrightarrow\sf{465=\cancel{\dfrac{15}{2}} \bigg[2a+(15-1)d\bigg]}\\\\\longrightarrow\sf{930=15[2a+14d]}\\\\\longrightarrow\sf{\cancel{930/15}=2a+14d}\\\\\longrightarrow\sf{62=2a+14d}\\\\\longrightarrow\sf{62=2(a+7d)}\\\\\longrightarrow\sf{\cancel{62/2} = a+7d}\\\\\longrightarrow\sf{31=a+7d.......................(2)}

    \underline{\boldsymbol{Using\:by\:Substitution\:method\::}}}

    From equation (2),we get;

    \longrightarrow\sf{a=31-7d.....................(3)}

    ∴Putting the value of a in equation (1),we get;

    \longrightarrow\sf{34=2(31-7d) +7d}\\\\\longrightarrow\sf{34=62 - 14d + 7d}\\\\\longrightarrow\sf{34=62-7d}\\\\\longrightarrow\sf{34-62=-7d}\\\\\longrightarrow\sf{-28=-7d}\\\\\longrightarrow\sf{d=\cancel{-28/-7}}\\\\\longrightarrow\bf{d=4}

    ∴Putting the value of d in equation (3),we get;

    \longrightarrow\sf{a=31-7(4)}\\\\\longrightarrow\sf{a=31-28}\\\\\longrightarrow\bf{a=3}

    Now;

    \longrightarrow\sf{S_{25}=\dfrac{25}{2} \bigg[2(3) + (25-1)(4) \bigg]}\\\\\\\longrightarrow\sf{S_{25} = \dfrac{25}{2} \bigg[6 + 24\times 4 \bigg]}\\\\\\\longrightarrow\sf{S_{25}=\dfrac{25}{2} \bigg[6+96\bigg]}\\\\\\\longrightarrow\sf{S_{25} = \dfrac{25}{\cancel{2}} \times \cancel{102}}\\\\\longrightarrow\sf{S_{25} = 25\times 51}\\\\\longrightarrow\bf{S_{25} = 1275}

    Thus;

    The sum of 25 terms will be 1275 .

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