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Ankita Raina to lead India's challenge in Fed Cup

Press Trust of India  |  Pune 

The country's top singles player, Ankita Raina, will lead the Indian challenge at Asia/Oceania Group I in Fed Cup, to be hosted in New Delhi's DLTA from February 7 to 10.

Raina, ranked 259 in the world, has been selected alongside the promising Karman Kaur Thandi (285), Pranjala Yadlapalli (482) and Prarthana Thombare, who is India's second-highest ranked player at number 144 after Sania Mirza (12).

Sania, who is carrying a knee injury and has already pulled out of the Australian Open, will not play in the event.

The selectors also have reserves in the talented Zeel Desai and Rutuja Bhosale, according to sources.

The All India Tennis Association (AITA) is expected to announce the team after getting clearance from the ITF. The selectors discussed the names recently after which AITA sent the names to the world governing body.

India will strive to qualify for the World Group II in the eight-team Group I.

The group features China, Chinese Taipei, Hong Kong, Japan, Kazakhstan, Korea and Thailand apart from the hosts.

There will be two pools of four team each and the winners of each pool will play off against one another to determine which nation advance to the World Group II play-offs, scheduled for April 2018.

The teams that finish third in each pool will play off against the nations finishing fourth in the other pool (A3 v B4 and B3 v A4). The two losers will be relegated to Asia/Oceania II in 2019.

(This story has not been edited by Business Standard staff and is auto-generated from a syndicated feed.)

First Published: Thu, January 04 2018. 17:30 IST