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Arithmetic Progressions Test - 25

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Arithmetic Progressions Test - 25
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  • Question 1
    1 / -0
    If $$\displaystyle l=20,d=-1,n=17$$ , then the first term is :
    Solution
    Given $$l = 20, d = -1, n = 17$$
    We need to find $$a$$.
    By using $$l=a+(n-1)d$$
    $$\therefore 20=a+(17-1)(-1)$$
    $$20=a-16$$
    $$a=20+16$$
    $$\therefore a=36$$
  • Question 2
    1 / -0
    $$50^{th}$$ term of the  AP. $$2, 5, 8, 11,.....$$ is :
    Solution
    Given series $$\displaystyle  2,5,8,11,.....$$ is in A.P.
    $$\therefore \displaystyle a=2,d=3$$
    We need to find $$50^{th}$$ term.
    $$\therefore n=50$$
    $$\therefore \displaystyle { a }_{ 50 }=a+49d$$
    $$=2+49\left( 3 \right) $$
    $$=2+147$$
    $$=149$$
  • Question 3
    1 / -0
    Find the sum of the following APs: $$-37, -33, -29, .....$$ to $$12$$ terms
    Solution
    Here, $$a=-37, d=4$$ and $$n=12$$
    Here we use the formula as given below:
    $$S_n = \dfrac{n}{2}[2a+(n-1)d]$$
    $$\therefore S_{12}=\dfrac {12}{2}(2\times -37+4\times 11)=6\times -30=-180$$
  • Question 4
    1 / -0
    Find the sum of the following APs: $$2, 7, 12, ......,$$ to $$10$$ terms
    Solution
    $$S=\dfrac {n}{2}(2a+d(n-1))$$

    Here, $$a=2, d=5$$ and $$n=10$$

    $$Sum=\dfrac {10}{2}(2(2)+45)$$

    $$=5\times 49=245$$
  • Question 5
    1 / -0
    In an AP, the $$9th$$ term is $$-72$$. The $$10th$$ term is $$60$$ less than the $$4th$$ term. Find the first term of the AP
    Solution
    Given that:
    $$a_9=-72$$ and $$a_{10}=a_4-60$$
    Let $$a$$ be the first term and $$d$$ be the common difference then
    $$a+8d=-72$$                            ......(1)
    And
    $$a+9d=a+3d-60$$
    $$\Rightarrow 6d=-60\Rightarrow d=-10.$$
    Substituting this value in equation (1), we get
    $$a+8\times (-10)=-72$$
    $$a=80-72$$
    $$a=8$$

    Hence,
    The first term is $$8.$$

  • Question 6
    1 / -0
    Find the sum of the following APs: $$0.6, 1.7, 2.8, ......$$ to $$100$$ terms
    Solution
    Here $$a=0.6, d=1.1$$ and $$n=100$$
    As we know, $$S_n = \dfrac {n}{2}[2a+(n-1)d]$$
    $$\therefore S_{100}=\dfrac {100}{2} (2(0.6)+1.1\times 99)=50\times 110.1=5505$$
  • Question 7
    1 / -0
    The first term of an AP is $$-50$$ and the $$50th$$ term is $$48$$. Find the common difference
    Solution
    Given,
    First term $$ a = -50 $$

    $$ 50^{th} $$ term, $$ T_{50} = 48 $$

    $$ d = ? $$

    $$ \boxed {T_{n} = a+(n-1)d} $$

    $$ T_{50} = 48 $$

    $$ 48 = -50+(50-1)d $$

    $$ 48 = -50+49d $$

    $$ 49d = 98 $$

    $$ \boxed {d = 2} $$ 
  • Question 8
    1 / -0
    The value obtained by subtracting the $$10^{th}$$ term of an AP from the $$17^{th}$$ term is $$56$$. Find the common difference.
    Solution
    In an Arithmetic Progression the general term of an A.P. is given by $$T_n$$:
    $$ \boxed {T_{n} = a+(n-1)d} $$
    According to the statement:
    $$ T_{17}-T_{10} = 56 $$
    $$ a+(17-1)d-[a+(10-1)d] = 56 $$
    $$ a+16d-[a+9d] = 56 $$
    $$ a+16d-a-9d = 56 $$
    $$ 7a = 56 $$
    $$\boxed {d = 8} $$
  • Question 9
    1 / -0
    The common difference of the sequence $$5,8,11,14,$$ is
    Solution
    Given, $$5,8,11,14$$
    $$\therefore 8-5=3$$
    $$11-8=3$$
    $$14-11=3$$
    The common difference of the sequence $$5,8,11,14$$ is $$3$$
  • Question 10
    1 / -0
    In an AP, the $$4th$$ term is $$36$$. The $$21st$$ term is $$108$$ more than the $$9th$$ term. Find the common difference.
    Solution
    Given: $$ T_{4} = 36 $$
    $$ T_{21} = 108+T_{9} $$
    $$ d = ? $$
    General term of an A.P. $$=T_{n} = a+(n-1)d $$
    $$T_{21} = 108+T_{9} $$
    $$ a+(21-1)d = 108+a+(9-1)d $$
    $$ 20d = 108 +8d $$
    $$ 12d = 108 $$
    $$ \boxed {d = 9} $$ 
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