## koli123able 2 years ago The sum of first 14 terms of 1,1.4,1.8,.... is equal to the sum of first n terms of 5.6,5.8,6.0,.... , find the value of n

1. sauravshakya

1,1.4,1.8,.... This is in AP with first term 1 and common difference 0.4

2. sauravshakya

Use sum formula of AP

3. amoodarya

14/2(a1+a14)=n/2(5.6+5.6+(n-1)*0.2)

4. amoodarya

14/2(1+1+13(.4))=n/2(5.6+5.6+(n-1)*0.2)

5. koli123able

i did the first A.P and its coming 50.4

6. sauravshakya

S14 =14/2 {2*1 + (14-1)0.4} = 50.4

7. koli123able

i am stuck here

8. sauravshakya

5.6,5.8,6.0,.... This is in AP with common difference 0.2 andfirt term 5.6

9. koli123able

$50.4=\frac{ n }{ 2 }[(2*5.6)+(n-1)*0.2]$

10. koli123able

stuck here

11. sauravshakya

Sn=50.4 n/2 { 2*5.6 + (n-1)0.2}=50.4 n{11 - 0.2n}=100.8

12. koli123able

2*5.6 its 11.2

13. sauravshakya

11n -0.2n^2 =100.8 11n -2/10 n^2 =100.8 110n -2n^2 =1008

14. sauravshakya

11.2-0.2=11

15. koli123able

oh

16. sauravshakya

110n -2n^2 =1008 2n^2-110n+1008=0 n^2-55n+504=0

17. koli123able

how did became 110n -2n^2 =1008

18. sauravshakya

|dw:1359028396466:dw|

19. koli123able

ok

20. sauravshakya

Now,n^2-55n+504=0 CAN u solve for n

21. koli123able

n= -11.61 and n= -43.38

22. koli123able

now what?

23. koli123able

24. sauravshakya

n=8 is surely not the correct solution

25. koli123able

the book says that answer is n=8

26. koli123able

how will it come 8

27. sauravshakya

Sorry I did a algebra mistake

28. sauravshakya

n=8 is correct

29. koli123able

how????? plz plz tell me

30. sauravshakya

Sn=50.4 n/2 { 2*5.6 + (n-1)0.2}=50.4 n/2 {11+0.2n}=50.4 n{11+0.2n}=100.8 11n+0,2n^2=100.8 11n +2/10 n^2 =100.8 110n+2n^2=1008 55n+n^2=504 n^2+55n-504=0 n^2+(63-8)n-504=0 n^2 +63n -8n-504=0 n(n+63)-8(n+63)=0 (n-8)(n+63)=0

31. sauravshakya

Now, either n-8=0 n=8 OR n+63=0 n=-63 BUT n cannot be negative So, n=8

32. sauravshakya

got it?

33. koli123able

thanks

34. sauravshakya

welcome