The first term of an AP is 10, the last term is 105 and the sum is 1150. Find the number of terms and the common difference.

Question

The first term of an AP is 10, the last term is 105 and the sum is 1150. Find the number of terms and the common difference.

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Margaret 3 weeks 2021-08-19T21:43:43+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-19T21:45:11+00:00

    \blue{\mathbb{\underline{ANSWER:}}}

    • The number of terms = 20.
    • Common difference = 5.

    \orange{\mathbb{\underline{GIVEN:}}}

    • The first term of an AP is 10
    • The last term is 105
    • Sum is 1150.

    \green{\mathbb{\underline{TO \ FIND:}}}

    • The number of terms.
    • Common difference.

    \purple{\mathbb{\underline{EXPLANATION:}}}

     \boxed{ \bold{S_n =\dfrac{n}{2}(a+l)}}

     \tt{S_n= 1150}

    a = 10

    l = 105

     \tt{1150 =\dfrac{n}{2}(10+ 105)}

    2300 = n(115)

    n = 20

     \boxed{ \bold{n = \dfrac{l - a}{d} + 1 }}

    n = 20

    l = 105

    a = 10

    \sf{20 = \dfrac{105 - 10}{d}+1}

    \sf{20 = \dfrac{95 + d}{d}}

    \sf{20d =95 + d}

    \sf{19d =95}

    \sf{d =5}

    Hence the common difference is 5 and number of terms is 20.

    \red{\mathbb{\underline{VERIFICATION:}}}

      \boxed{\bold{S_n =  \dfrac{n}{2} (2a + (n-1)d)}}

    n = 20

    a = 10

    d = 5

     \sf{S_n =  \dfrac{20}{2} (2(10) + (20-1)5)}

     \sf{S_n =  10 (20+ 19(5))}

    \sf {S_n =  10 (20+95)}

     \sf{S_n =  10 (115)}

     \sf{S_n =  1150}

    HENCE VERIFIED.

    0
    2021-08-19T21:45:40+00:00

    Given ,

    First term (a) = 10

    nth term (an) = 105

    Sum of first n terms (Sn) = 1150

    We know that , the sum of first n terms of an AP is given by

     \boxed{  \sf{S_{n} =  \frac{n}{2} (a +  a_{n} )}}

    Thus ,

    1150 = n/2 × (10 + 105)

    2300 = 115n

    n = 2300/115

    n = 20

    Since , nth term (an) = 105

    Thus ,

    105 = 10 + (20 – 1)d

    95 = 19d

    d = 95/19

    d = 5

    Therefore ,

    • The number of terms and the common difference of given AP are 20 and 5

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