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Let first term be 'a' and common difference be 'd' of A.P. <br> `therefore" "T_(p)=a+(p-1)d=(1)/(q)" …(1)"` <br> `T_(q)=a(q-1)d=(1)/(p)" …(2)"` <br> Subtracting equation (2) from equation (1) <br> `rArr" "(p-q)d=(1)/(q)-(1)/(p)=(p-q)/(pq)" "rArr d=(1)/(pq)` <br> Put the value of d in equation (1), <br> `a+(p-1)cdot(1)/(pq)=(1)/(q)" "rArr a=(1)/(pq)` <br> Now, sum of pq terms <br> `S_(pq)=(pq)/(2)[2cdot(1)/(pq)+(pq-1)(1)/(pq)]` <br> `=(1)/(2)(pq+1)`**Representation of sequences and different types of series**

**Definition + algorithm to determine the sequence of AP**

**General term of an AP**

**`n^(th)` term of an AP from the end**

**The `n^(th)` term of the sequence is `3n-2` . Is the sequence an AP. If so; find the 10th term .**

**If first term is 8 and last term is 20 common diffference is 2 . find the value of n when the series are in AP.**

**Which term of the sequence is the first negative term .. `20; 19(1/4);18(1/2);17(3/4).....`**

**Show that the sum of `(m+n)^(th) and (m-n)^(th)` term of an AP is equal to twice the `m^(th)` term ?**

**Show that sum `S_n` of n terms of an AP with first term a and common difference d is `S_n=n/2(2a+(n-1)d)`**

**Find the sum of 20 terms of the AP 1, 4, 7, 10,....**