Question

the first term of an ap is -1 and 350 is the last term if the common difference is .find the 10 th term​

Answer 1

Answer:

Let a and d are first term and common difference for an AP.

Let a and d are first term and common difference for an AP.Number of terms of AP is n

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=an

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1∴n=38

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1∴n=38Therefore, number of terms in given AP  is n=38

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1∴n=38Therefore, number of terms in given AP  is n=38Sum of n terms of AP is Sn

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1∴n=38Therefore, number of terms in given AP  is n=38Sum of n terms of AP is SnSn=2n(a+l)

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1∴n=38Therefore, number of terms in given AP  is n=38Sum of n terms of AP is SnSn=2n(a+l)here n=38

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1∴n=38Therefore, number of terms in given AP  is n=38Sum of n terms of AP is SnSn=2n(a+l)here n=38S38=238[17+350]

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1∴n=38Therefore, number of terms in given AP  is n=38Sum of n terms of AP is SnSn=2n(a+l)here n=38S38=238[17+350]       =19×367

Let a and d are first term and common difference for an AP.Number of terms of AP is nLast term =nth term =l=anWe have a=17,d=9,l=350We know a+(n−1)d=350⇒17+(n−1)9=350⇒(n−1)9=350−17⇒(n−1)9=333⇒n−1=333/9⇒n−1=37⇒n=37+1∴n=38Therefore, number of terms in given AP  is n=38Sum of n terms of AP is SnSn=2n(a+l)here n=38S38=238[17+350]       =19×367S38=6973