**Category : **10th Class

A sequence is said to be in A.P, if the difference between the consecutive terms is a constant. The difference between the consecutive terms of an AP is called common difference and any general term is called nth term of the sequence.

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},---,{{a}_{n}}\] be the nth terms of the sequence in A.P., then nth terms of the sequence is given by \[{{a}_{n}}=a+\left( n-1 \right)d\],

Where 'a' is the first term of the sequence

'd' is the common difference

'n' is the number of terms of the sequence.

**Sum of N-Terms of the A.P.**

If \[{{a}_{1}},{{a}_{2}},{{a}_{3}},---,{{a}_{n}}\] be the n terms of the sequence in A.P., then the sum of n-terms of the sequence is given \[{{S}_{n}}=\frac{n}{2}\left[ 2a+(n-1)d \right]\]

**Arithmetic Mean**

If 'a' and 'b' are any two terms of A.P., then the arithmetic mean is given by \[AM=\frac{a+b}{2}\]

**Properties of AP**

(a) If a constant is added or subtracted from each term of the AP sequence, then the resulting sequence is also in AP.

(b) If each term of an AP sequence is multiplied or divided by a constant, then the resulting sequence is also in AP.

** For a sequence in A.P., the sum of n terms of the sequence is \[2n+3{{n}^{2}}\], find the 20th term of the sequence:**

(a) 121

(b) 112

(c) 119

(d) 124

(e) None of these

**Answer:** (c)

**Explanation**

We have, \[{{S}_{n}}=2n+3{{n}^{2}}\]

\[{{S}_{1}}=5,{{S}_{2}}=16\]

Thus common difference is \[d={{S}_{2}}-2a=16-10=6\]

Hence the 20th term of the sequence is \[=6\times 320-1=119\]

**Find the number of odd integers starting with 15 and will give the sum as 975.**

(a) 25

(b) 15

(c) 20

(d) 30

(e) None of these

**Answer:** (a)

**Explanation**

Here we have a = 15 and common difference = 2

Since we know that, \[{{S}_{n}}=\frac{n}{2}\left[ 2a+(n-1)d \right]\]

But \[{{S}_{n}}=975\]

\[\therefore \,\,\,\frac{n}{2}\left[ 2a+(n-1)d \right]=975\]

\[\Rightarrow \,\,\frac{n}{2}\left[ 2\times 15+(n-1)\times 2 \right]=975\]

\[\Rightarrow \,\,{{n}^{2}}+14n-=975=0\]

\[\Rightarrow \,\,\,\,\,\,n=25\,\,\text{or}\,\,n=-39\]

Since number of terms cannot be negative, hence n = - 39 is rejected. Therefore n = 25.

**If the 15th term of an AP is 45 and 20th term is 60, and then find the 30th term of the AP.**

(a) 70

(b) 90

(c) 110

(d) 120

(e) None of these

**Answer:** (b)

**Explanation**

We have,

\[{{A}_{15}}=A+14d=45\]

\[{{a}_{20}}=a+19d=60\]

On solving the above equation we get,

a = 3 and d = 3

Therefore, \[{{a}_{30}}=a+29d\]

\[\Rightarrow \,\,{{a}_{30}}=3+29\times 3=90\]

**If the number of different cards of different colours Thomas has are in AP. If he has cards of seven different colours in the order of VIBGYOR such that third colour is four times the first colours and the sixth colours is 17, then the number of fifth colour cards is:**

(a) 14

(b) 16

(c) 12

(d) 10

(e) None of these

**Answer:** (a)

** The 10th common term between the series \[3+7+11+---and\,1+6+11+---\] is given by:**

(a) 175

(b) 155

(c) 187

(d) 191

(e) None of these

**Answer:** (d)

*play_arrow*Introduction*play_arrow*Arithmetic Progression*play_arrow*Geometric Progression*play_arrow*Harmonic Progression*play_arrow*Sequence and Series

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