Q. 9

# The 17^{th} term of an AP exceeds its 10^{th} term by 7. Find the common difference.

Answer :

Sum of the first n term is given by (s_{n}) = [2a + (n-1) d]

The sum of first 7 terms S_{7} = 49 → 7^{2}

S_{7} = [2×a+(7-1)d]

49 = [2a+6d]

49 = 7[a+3d]

7 = a + 3d ⇒ __1__

The sum of first 17 terms S_{17} = 289 → 17^{2}

S_{17} = [2×a+(17-1)d]

289 = [2a+16d]

289 = 17[a+8d]

17 = a + 8d ⇒ __2__

By subtracting both the equations

17 – 7 = a + 8d – (a+3d)

10 = 5d

d =2

∴ common difference (d) = 2

Substituting‘d’ in equation __1__

7 = a + 3d

7 = a + 3(2)

a = 1

∴ Sum of the first n term is given by (s_{n}) = [2a + (n-1) d]

= [2×1 + (n-1)2]

= n[ 1+ (n-1)]

= n^{2}

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Show that a_{1}, a_{2},…..a_{n} form an AP where is defined as below :

a_{n} = 9-5n

Also find the sum of the first 15 terms in each case.

AP- MathematicsShow that a_{1}, a_{2},…..a_{n} form an AP where is defined as below :

a_{n} = 3+4n

Also find the sum of the first 15 terms in each case.

AP- Mathematics