Math, asked by havvagb786, 9 months ago

find the sum of all the terms of the arthimetic squance 7,13,19,...,61​

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Answered by dhruv220605
1

Answer:

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❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)d

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore,

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6205-7=(n-1)6

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6205-7=(n-1)6198=(n-1)6

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6205-7=(n-1)6198=(n-1)6198/6=(n-1)

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6205-7=(n-1)6198=(n-1)6198/6=(n-1)33=(n-1)

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6205-7=(n-1)6198=(n-1)6198/6=(n-1)33=(n-1)n=33+1=34

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6205-7=(n-1)6198=(n-1)6198/6=(n-1)33=(n-1)n=33+1=34•°• there are 34 terms in the given Arithmetic progression 7,13,19.....205.

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6205-7=(n-1)6198=(n-1)6198/6=(n-1)33=(n-1)n=33+1=34•°• there are 34 terms in the given Arithmetic progression 7,13,19.....205.☘️THANK YOU!☘️

❄️❄️❄️❄️❄️❄️❄️❄️❄️❄️let a be the 1st term.Therefore, a=7.2nd term =a+d 7+d=13d=13-7=6d=6.aⁿ=a+(n-1)daⁿ=7+(n-1)6w.k.t,aⁿ=205therefore, 205=7+(n-1)6205-7=(n-1)6198=(n-1)6198/6=(n-1)33=(n-1)n=33+1=34•°• there are 34 terms in the given Arithmetic progression 7,13,19.....205.☘️THANK YOU!☘️ALL THE BEST..☺️

Answered by tennetiraj86
0

Answer:

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