If 7 times the 7th term of an A.P. is equal to 11 times the 11th term,show that 18th term of A.P. is 0.

Let 1st term of AP be 'a',

Let the common difference be 'd',

So,as per question, 7 x 7th term = 11 x 11th term

or, 7 [a+6d]  = 11 [a+10d]

or, 7a+42d = 11a +110d

or, 4a + 68d = 0

or, a + 17d = 0 .......(1)

Now, 18th term = a+17d

or, t18 = 0   { From (1)}

                                            Proved

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7A7 = 11 A11 

 7(a+ 6d ) =11(a+ 10d)

7a + 42d = 11a + 110d

 4a + 68d= 0

4(a+ 17d)= 0

 a+ 17d =0

hence A18 = 0   PROVED

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7th term?7=11th term ?11
=>( a+(7-1)d)7=11(a+(11-1)d)
=>7a+42d=11a+110d
=>4a+68d=0
=>a+17d=0
=>a+(18-1)d
=>18th term =0
Proved
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7 times 7 the term of AP

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Let 'a' be the first term and 'd' be the common difference of the given AP.

           7th term of the AP = a + (7-1)d= a+6d
​          11th term of the AP = a = (11-1)d=a+10d
​           18th term of the AP= a +(18-1)d = a +17d      
​             7(a+6d)=11(a+10d)
             7a+42d=11a+110d
​             11a-7a+110d-42d=0
             4a+68d=0
             4(a+17d)=0
               a +17d=0
     Therefore 18th term of AP is equal to zero.
                
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7A7 = 11 A11 7(a+ 6d ) =11(a+ 10d) 7a + 42d = 11a + 110d 4a + 68d= 0 4(a+ 17d)= 0 a+ 17d =0 hence A18 = 0 PROVED
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