The sum of the first five terms of an arithmetic sequence is 150 and the sum of the first ten terms is 550. (b) What is the eighth ter

Question

The sum of the first five terms of an arithmetic sequence is 150 and the sum of the first ten terms is 550.
(b) What is the eighth term?

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4 days 2021-10-11T13:19:37+00:00 2 Answers 0 views 0

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    0
    2021-10-11T13:20:54+00:00

    The sum of the first 10 terms of the AP is – 150 and the sum of the first 20 terms is 0150–550 = -700.

    S10 = (10/2)[2a+(10–1)d] = -150, or

    -30 = 2a + 9d, or

    2a = -30–9d …(1)

    S20 = -700 = (20/2)[2a+(20–1)d], or

    -70 = 2a + 19d, or

    2a = -70–19d …(2)

    Equate (1) and (2),

    -30–9d = -70–19d, or

    10d = -40, or

    d = -4.

    From (1), 2a = -30+36 = 6, or a = 3.

    The first term of the AP = 3 and the common ratio = -4. Or the AP is 3, -1, -5, -9,…

    0
    2021-10-11T13:21:18+00:00

    SOLUTION :

    \bigstar Sum of the first five term of an arithmetic sequence is 150.

    Using formula :

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

    A/q

    \longrightarrow\sf{\cancel{150}=\dfrac{\cancel{5}}{2} \bigg[2a+(5-1)d\bigg]}\\\\\\\longrightarrow\sf{30\times 2=2a+4d}\\\\\\\longrightarrow\sf{60=2a+4d....................(1)}

    Again,

    \bigstar Sum of the first ten term of an arithmetic sequence is 550.

    \longrightarrow\sf{\cancel{550}=\dfrac{\cancel{10}}{2} \bigg[2a+(10-1)d\bigg]}\\\\\\\longrightarrow\sf{55\times 2=2a+9d}\\\\\\\longrightarrow\sf{110=2a+9d....................(2)}

    ∴ Substituting equation (1) & equation (2),we get;

    From eq. (1) :

    \longrightarrow\sf{60=2a+4d}\\\\\longrightarrow\sf{2a=60-4d}\\\\\longrightarrow\bf{a=60-4d/2......................(3)}

    ∴ Putting the value of a in equation (2),we get;

    \longrightarrow\sf{110=2\bigg(\dfrac{60-4d}{2} \bigg)+9d}\\\\\\\longrightarrow\sf{110=\dfrac{120-8d}{2} +9d}\\\\\\\longrightarrow\sf{220=120-8d+18}\\\\\\\longrightarrow\sf{220=120+10d}\\\\\\\longrightarrow\sf{10d=220-120}\\\\\\\longrightarrow\sf{10d=100}\\\\\\\longrightarrow\sf{d=\cancel{100/10}}\\\\\\\longrightarrow\bf{d=10}

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

    \longrightarrow\sf{a=\dfrac{60-4(10)}{2} }\\\\\\\longrightarrow\sf{a=\dfrac{60-40}{2} }\\\\\\\longrightarrow\sf{a=\cancel{\dfrac{20}{2} }}\\\\\\\longrightarrow\bf{a=10}

    Now,

    \longrightarrow\sf{S_8=\dfrac{n}{2} \bigg[2a+(n-1)d\bigg]}\\\\\\\longrightarrow\sf{S_8=\cancel{\dfrac{8}{2}} \bigg[2(10)+(8-1)10\bigg]}\\\\\\\longrightarrow\sf{S_8=4[20+7\times 10]}\\\\\\\longrightarrow\sf{S_8=4[20+70]}\\\\\\\longrightarrow\sf{S_8=4[90]}\\\\\\\longrightarrow\sf{S_8=4\times 90}\\\\\\\longrightarrow\bf{S_8=360}

    Thus;

    The eight term will be 360 .

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