SolveItClass 10 · NCERT

NCERT Solutions · Class 10 Mathematics Arithmetic Progressions

49 questions · 49 still being checked

EXERCISE 5.3 1–10 (part 4 of 6)

  1. Exercise 1

    Find the sum of the following APs:
    (i)
    2\displaystyle 2, 7\displaystyle 7, 12\displaystyle 12, . . ., to 10\displaystyle 10 terms.
    (ii)
    -37\displaystyle 37, -33\displaystyle 33, -29\displaystyle 29, . . , to 12\displaystyle 12 terms.
    (iii)
    0.6\displaystyle 6, 1.7\displaystyle 1.7, 2.8\displaystyle 2.8, . . ,, to 100\displaystyle 100 terms.
    (iv)
    115,112,110,…\displaystyle \frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots, to 11\displaystyle 11 terms.
    NCERT’s answer
    (i)
    $\displaystyle 245$ (ii) -$\displaystyle 180$ (iii) $\displaystyle 5505$ (iv) $\displaystyle \frac{33}{20}$

    Working being prepared

  2. Exercise 2

    Find the sums given below :
    (i)
    7+1012+14+…+84\displaystyle 7+10 \frac{1}{2}+14+\ldots+84
    (ii)
    34+32+30+…+10\displaystyle 34+32+30+\ldots+10
    (iii)
    −5+(−8)+(−11)+…+(−230)\displaystyle -5+(-8)+(-11)+\ldots+(-230)
    NCERT’s answer
    (i)
    $\displaystyle 1046 \frac{1}{2}$ (ii) $\displaystyle 286$ (iii) -$\displaystyle 8930$

    Working being prepared

  3. Exercise 3

    In an AP:
    (i)
    given a=5,d=3,an=50\displaystyle a=5, d=3, a_{n}=50, find n\displaystyle n and Sn\displaystyle \mathrm{S}_{n}.
    (ii)
    given a=7,a13=35\displaystyle a=7, a_{13}=35, find d\displaystyle d and S13\displaystyle \mathrm{S}_{13}.
    (iii)
    given a12=37,d=3\displaystyle a_{12}=37, d=3, find a\displaystyle a and S12\displaystyle \mathrm{S}_{12}.
    (iv)
    given a3=15, S10=125\displaystyle a_{3}=15, \mathrm{~S}_{10}=125, find d\displaystyle d and a10\displaystyle a_{10}.
    (v)
    given d=5, S9=75\displaystyle d=5, \mathrm{~S}_{9}=75, find a\displaystyle a and a9\displaystyle a_{9}.
    (vi)
    given a=2,d=8, Sn=90\displaystyle a=2, d=8, \mathrm{~S}_{n}=90, find n\displaystyle n and an\displaystyle a_{n}.
    (vii)
    given a=8,an=62, Sn=210\displaystyle a=8, a_{n}=62, \mathrm{~S}_{n}=210, find n\displaystyle n and d\displaystyle d.
    (viii)
    given an=4,d=2, Sn=−14\displaystyle a_{n}=4, d=2, \mathrm{~S}_{n}=-14, find n\displaystyle n and a\displaystyle a.
    (ix)
    given a=3,n=8, S=192\displaystyle a=3, n=8, \mathrm{~S}=192, find d\displaystyle d.
    (x)
    given l=28, S=144\displaystyle l=28, \mathrm{~S}=144, and there are total 9\displaystyle 9 terms. Find a\displaystyle a.
    NCERT’s answer
    (i)
    $\displaystyle n=16, \mathrm{~S}_{n}=440$ (ii) $\displaystyle d=\frac{7}{3}, \mathrm{~S}_{13}=273$ (iii) $\displaystyle a=4, \mathrm{~S}_{12}=246$ (iv) $\displaystyle d=-1, a_{10}=8$ (v) $\displaystyle a=-\frac{35}{3}, a_{9}=\frac{85}{3}$ (vi) $\displaystyle n=5, a_{n}=34$ (vii) $\displaystyle n=6, d=\frac{54}{5}$ (viii) $\displaystyle n=7, a=-8$ (ix) $\displaystyle d=6$ (x) $\displaystyle a=4$

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  4. Exercise 4

    How many terms of the AP : 9\displaystyle 9, 17\displaystyle 17, 25\displaystyle 25, . . . must be taken to give a sum of 636\displaystyle 636?
    NCERT’s answer
    12. By putting $\displaystyle a=9, d=8, \mathrm{~S}=636$ in the formula $\displaystyle \mathrm{S}=\frac{n}{2}[2 a+(n-1) d]$, we get a quadratic equation $\displaystyle 4 n^{2}+5 n-636=0$. On solving, we get $\displaystyle n=-\frac{53}{4}, 12$. Out of these two roots only one root $\displaystyle 12$ is admissible.

    Working being prepared

  5. Exercise 5

    The first term of an AP is 5\displaystyle 5, the last term is 45\displaystyle 45 and the sum is 400. Find the number of terms and the common difference.
    NCERT’s answer
    $\displaystyle n=16, d=\frac{8}{3}$

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  6. Exercise 6

    The first and the last terms of an AP are 17\displaystyle 17 and 350\displaystyle 350 respectively. If the common difference is 9\displaystyle 9, how many terms are there and what is their sum?
    NCERT’s answer
    $\displaystyle n=38, \mathrm{~S}=6973$

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  7. Exercise 7

    Find the sum of first 22\displaystyle 22 terms of an AP in which d=7\displaystyle d=7 and 22nd term is 149.
    NCERT’s answer
    Sum $\displaystyle =1661$

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  8. Exercise 8

    Find the sum of first 51\displaystyle 51 terms of an AP whose second and third terms are 14\displaystyle 14 and 18\displaystyle 18 respectively.
    NCERT’s answer
    $\displaystyle \mathrm{S}_{51}=5610$

    Working being prepared

  9. Exercise 9

    If the sum of first 7\displaystyle 7 terms of an AP is 49\displaystyle 49 and that of 17\displaystyle 17 terms is 289\displaystyle 289 , find the sum of first n\displaystyle n terms.
    NCERT’s answer
    $\displaystyle n^{2}$

    Working being prepared

  10. Exercise 10

    Show that a1,a2,…,an,…\displaystyle a_{1}, a_{2}, \ldots, a_{n}, \ldots form an AP where an\displaystyle a_{n} is defined as below :
    (i)
    an=3+4n\displaystyle a_{n}=3+4 n
    (ii)
    an=9−5n\displaystyle a_{n}=9-5 n Also find the sum of the first 15\displaystyle 15 terms in each case.
    NCERT’s answer
    (i)
    $\displaystyle \mathrm{S}_{15}=525$ (ii) $\displaystyle \mathrm{S}_{15}=-465$

    Working being prepared